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a tiny misprint corrected
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Wadim Zudilin
Wadim Zudilin
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Dear Ross,

It looks that you don't really wish to see known formulae for your zigzag numbers. Otherwise I don't understand why you found my search insufficient.

The OEIS A000111 gives the formula $$ A_m=2^n\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$$$ A_m=2^m\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ where $E_m(x)$ are the Euler polynomials which can be generated by the following explicit expansion $$ E_m(x)=\sum_{n=0}^m\frac1{2^n}\sum_{k=0}^n(-1)^k\binom nk(x+k)^m, $$ a double sum as in your case. Even if this formula is not exactly the same as yours (although it looks pretty similar), this is a known double sum expression for $A_m$. There is a lot of room for playing with this double sum and producing many other (useful and useless) formulae for the zigzag numbers.

And don't forget: I've never seen this specific sequence before.

Best wishes, Wadim

Dear Ross,

It looks that you don't really wish to see known formulae for your zigzag numbers. Otherwise I don't understand why you found my search insufficient.

The OEIS A000111 gives the formula $$ A_m=2^n\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ where $E_m(x)$ are the Euler polynomials which can be generated by the following explicit expansion $$ E_m(x)=\sum_{n=0}^m\frac1{2^n}\sum_{k=0}^n(-1)^k\binom nk(x+k)^m, $$ a double sum as in your case. Even if this formula is not exactly the same as yours (although it looks pretty similar), this is a known double sum expression for $A_m$. There is a lot of room for playing with this double sum and producing many other (useful and useless) formulae for the zigzag numbers.

And don't forget: I've never seen this specific sequence before.

Best wishes, Wadim

Dear Ross,

It looks that you don't really wish to see known formulae for your zigzag numbers. Otherwise I don't understand why you found my search insufficient.

The OEIS A000111 gives the formula $$ A_m=2^m\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ where $E_m(x)$ are the Euler polynomials which can be generated by the following explicit expansion $$ E_m(x)=\sum_{n=0}^m\frac1{2^n}\sum_{k=0}^n(-1)^k\binom nk(x+k)^m, $$ a double sum as in your case. Even if this formula is not exactly the same as yours (although it looks pretty similar), this is a known double sum expression for $A_m$. There is a lot of room for playing with this double sum and producing many other (useful and useless) formulae for the zigzag numbers.

And don't forget: I've never seen this specific sequence before.

Best wishes, Wadim

Toss to Ross, correct unless known nickname
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Will Jagy
Will Jagy
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Dear TossRoss,

It looks that you don't really wish to see known formulae for your zigzag numbers. Otherwise I don't understand why you found my search insufficient.

The OEIS A000111 gives the formula $$ A_m=2^n\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ where $E_m(x)$ are the Euler polynomials which can be generated by the following explicit expansion $$ E_m(x)=\sum_{n=0}^m\frac1{2^n}\sum_{k=0}^n(-1)^k\binom nk(x+k)^m, $$ a double sum as in your case. Even if this formula is not exactly the same as yours (although it looks pretty similar), this is a known double sum expression for $A_m$. There is a lot of room for playing with this double sum and producing many other (useful and useless) formulae for the zigzag numbers.

And don't forget: I've never seen this specific sequence before.

Best wishes, Wadim

Dear Toss,

It looks that you don't really wish to see known formulae for your zigzag numbers. Otherwise I don't understand why you found my search insufficient.

The OEIS A000111 gives the formula $$ A_m=2^n\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ where $E_m(x)$ are the Euler polynomials which can be generated by the following explicit expansion $$ E_m(x)=\sum_{n=0}^m\frac1{2^n}\sum_{k=0}^n(-1)^k\binom nk(x+k)^m, $$ a double sum as in your case. Even if this formula is not exactly the same as yours (although it looks pretty similar), this is a known double sum expression for $A_m$. There is a lot of room for playing with this double sum and producing many other (useful and useless) formulae for the zigzag numbers.

And don't forget: I've never seen this specific sequence before.

Best wishes, Wadim

Dear Ross,

It looks that you don't really wish to see known formulae for your zigzag numbers. Otherwise I don't understand why you found my search insufficient.

The OEIS A000111 gives the formula $$ A_m=2^n\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ where $E_m(x)$ are the Euler polynomials which can be generated by the following explicit expansion $$ E_m(x)=\sum_{n=0}^m\frac1{2^n}\sum_{k=0}^n(-1)^k\binom nk(x+k)^m, $$ a double sum as in your case. Even if this formula is not exactly the same as yours (although it looks pretty similar), this is a known double sum expression for $A_m$. There is a lot of room for playing with this double sum and producing many other (useful and useless) formulae for the zigzag numbers.

And don't forget: I've never seen this specific sequence before.

Best wishes, Wadim

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Wadim Zudilin
Wadim Zudilin
  • 13.5k
  • 4
  • 58
  • 102

Dear Toss,

It looks that you don't really wish to see known formulae for your zigzag numbers. Otherwise I don't understand why you found my search insufficient.

The OEIS A000111 gives the formula $$ A_m=2^n\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ where $E_m(x)$ are the Euler polynomials which can be generated by the following explicit expansion $$ E_m(x)=\sum_{n=0}^m\frac1{2^n}\sum_{k=0}^n(-1)^k\binom nk(x+k)^m, $$ a double sum as in your case. Even if this formula is not exactly the same as yours (although it looks pretty similar), this is a known double sum expression for $A_m$. There is a lot of room for playing with this double sum and producing many other (useful and useless) formulae for the zigzag numbers.

And don't forget: I've never seen this specific sequence before.

Best wishes, Wadim