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Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$.

The sequence begins with $$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704, 8555388, 72442465, 647479819$$ $a(n)$ is given by the following recurrence relation, due to David Callan $$a(n)=\sum\limits_{k=1}^{n}u(n,k) ,$$ where $$u(n,k)=u(n-1,k-1)+k\sum\limits_{j=k}^{n-1}u(n-1,j), u(n,0)=[n=0] .$$ Here square brackets denote Iverson brackets.

Let $b(n)$ be a sequence of the positive integers such that $$b(2^m(2n+1))=\sum\limits_{k=0}^{m}(k+1)b(2^k n), \qquad b(0)=1 .$$

The sequence begins with $$1, 1, 3, 1, 6, 3, 7, 1, 10, 6, 15, 3, 25, 7, 15, 1, 15, 10, 26, 6, 45, 15, 33, 3, 65$$

Let $s(n)$ be a sequence of the positive integers such that $$s(n)=\sum\limits_{k=0}^{2^n-1}b(k)$$

I conjecture that $$s(n)=a(n+1)$$

Is there a way to prove it?

Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$.

The sequence begins with $$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704, 8555388, 72442465, 647479819$$ $a(n)$ is given by the following recurrence relation, due to David Callan $$a(n)=\sum\limits_{k=1}^{n}u(n,k),$$ where $$u(n,k)=u(n-1,k-1)+k\sum\limits_{j=k}^{n-1}u(n-1,j), \qquad u(n,0)=[n=0].$$ Here square brackets denote Iverson brackets.

Let $b(n)$ be a sequence of the positive integers such that $$b(2^m(2n+1))=\sum\limits_{k=0}^{m}(k+1)b(2^k n), \qquad b(0)=1 .$$

The sequence begins with $$1, 1, 3, 1, 6, 3, 7, 1, 10, 6, 15, 3, 25, 7, 15, 1, 15, 10, 26, 6, 45, 15, 33, 3, 65$$

Let $s(n)$ be a sequence of the positive integers such that $$s(n)=\sum\limits_{k=0}^{2^n-1}b(k)$$

I conjecture that $$s(n)=a(n+1)$$

Is there a way to prove it?

Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$.

The sequence begins with $$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704, 8555388, 72442465, 647479819$$ $a(n) is given by the following recurrence relation, due to David Callan $$a(n)=\sum\limits_{k=1}^{n}u(n,k) ,$$ where $$u(n,k)=u(n-1,k-1)+k\sum\limits_{j=k}^{n-1}u(n-1,j), u(n,0)=[n=0] .$$ Here square brackets denote Iverson brackets.

Let $b(n)$ be a sequence of the positive integers such that $$b(2^m(2n+1))=\sum\limits_{k=0}^{m}(k+1)b(2^k n), \qquad b(0)=1 .$$

The sequence begins with $$1, 1, 3, 1, 6, 3, 7, 1, 10, 6, 15, 3, 25, 7, 15, 1, 15, 10, 26, 6, 45, 15, 33, 3, 65$$

Let $s(n)$ be a sequence of the positive integers such that $$s(n)=\sum\limits_{k=0}^{2^n-1}b(k)$$

I conjecture that $$s(n)=a(n+1)$$

Is there a way to prove it?

Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$.

The sequence begins with $$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704, 8555388, 72442465, 647479819$$ $a(n)$ is given by the following recurrence relation, due to David Callan $$a(n)=\sum\limits_{k=1}^{n}u(n,k) ,$$ where $$u(n,k)=u(n-1,k-1)+k\sum\limits_{j=k}^{n-1}u(n-1,j), u(n,0)=[n=0] .$$ Here square brackets denote Iverson brackets.

Let $b(n)$ be a sequence of the positive integers such that $$b(2^m(2n+1))=\sum\limits_{k=0}^{m}(k+1)b(2^k n), \qquad b(0)=1 .$$

The sequence begins with $$1, 1, 3, 1, 6, 3, 7, 1, 10, 6, 15, 3, 25, 7, 15, 1, 15, 10, 26, 6, 45, 15, 33, 3, 65$$

Let $s(n)$ be a sequence of the positive integers such that $$s(n)=\sum\limits_{k=0}^{2^n-1}b(k)$$

I conjecture that $$s(n)=a(n+1)$$

Is there a way to prove it?

Let $a(n)$ be A113227, i.e., the number of permutations avoiding the pattern $1-23-4$.

The sequence begins with $$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704, 8555388, 72442465, 647479819$$ Here (recurrence due to David Callan) $$a(n)=\sum\limits_{k=1}^{n}u(n,k)$$ where $$u(n,k)=u(n-1,k-1)+k\sum\limits_{j=k}^{n-1}u(n-1,j), u(n,0)=[n=0]$$ Here square brackets denote Iverson brackets.

Let $b(n)$ be a sequence of the positive integers such that $$b(2^m(2n+1))=\sum\limits_{k=0}^{m}(k+1)b(2^k n), b(0)=1$$

The sequence begins with $$1, 1, 3, 1, 6, 3, 7, 1, 10, 6, 15, 3, 25, 7, 15, 1, 15, 10, 26, 6, 45, 15, 33, 3, 65$$

Let $s(n)$ be a sequence of the positive integers such that $$s(n)=\sum\limits_{k=0}^{2^n-1}b(k)$$

I conjecture that $$s(n)=a(n+1)$$

Is there a way to prove it?

Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$.

The sequence begins with $$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704, 8555388, 72442465, 647479819$$ $a(n) is given by the following recurrence relation, due to David Callan $$a(n)=\sum\limits_{k=1}^{n}u(n,k) ,$$ where $$u(n,k)=u(n-1,k-1)+k\sum\limits_{j=k}^{n-1}u(n-1,j), u(n,0)=[n=0] .$$ Here square brackets denote Iverson brackets.

Let $b(n)$ be a sequence of the positive integers such that $$b(2^m(2n+1))=\sum\limits_{k=0}^{m}(k+1)b(2^k n), \qquad b(0)=1 .$$

The sequence begins with $$1, 1, 3, 1, 6, 3, 7, 1, 10, 6, 15, 3, 25, 7, 15, 1, 15, 10, 26, 6, 45, 15, 33, 3, 65$$

Let $s(n)$ be a sequence of the positive integers such that $$s(n)=\sum\limits_{k=0}^{2^n-1}b(k)$$

I conjecture that $$s(n)=a(n+1)$$

Is there a way to prove it?