

A152537


Convolution sequence: convolved with A000041 = powers of 2, (A000079).


5



1, 1, 1, 2, 4, 9, 18, 37, 74, 148, 296, 592, 1183, 2366, 4732, 9463, 18926, 37852, 75704, 151408, 302816, 605632, 1211265, 2422530, 4845060, 9690120, 19380241, 38760482, 77520964, 155041928, 310083856, 620167712, 1240335424, 2480670848, 4961341696, 9922683391
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OFFSET

0,4


COMMENTS



LINKS



FORMULA

Construct an array of rows such that nth row = partial sums of (n1)th row
of A010815: (1, 1, 1, 0, 0, 1, 0, 1,...).
A152537 = sums of antidiagonal terms of the array.
The sequence may be obtained directly from the following set of operations:
Our given sequence = A000041: (1, 1, 2, 3, 5, 7, 11,...). Delete the first
"1" then consider (1, 2, 3, 5, 7, 11,...) as an operator Q which we write in reverse with 1,2,3,...terms for each operation. Letting R = the target sequence (1,2,4,8,...); we begin a(0) = 1, a(1) = 1, then perform successive
operations of: "next term in (1,2,4,...)  dot product of Q*R" where Q is
written right to left and R (the ongoing result) written left to right).
Examples: Given 4 terms Q, R, we have: (5,3,2,1) dot (1,1,1,2) = (5+3+2+2) =
12, which we subtract from 16, = 4.
Given 5 terms of Q,R and A152537, we have (7,5,3,2,1) dot (1,1,1,2,4) = 23
which is subtracted from 32 giving 9. Continue with analogous operations to generate the series.


EXAMPLE

a(5) = 9 = 32  23 = (32  ((7,5,3,2,1) dot (1,1,1,2,4)))
(1,1,2,3) convolved with (1,1,1,2) = 8, where (1,1,2,3...) = the first four partition numbers.


MATHEMATICA

nmax = 40; CoefficientList[Series[Product[1x^k, {k, 1, nmax}] / (12*x), {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 02 2018 *)


PROG

(PARI) /* computation by definition (division of power series) */
N=55;
A000041=vector(N, n, numbpart(n1));
S152537=S000079/S000041;
(PARI) /* computation using power series eta(x) and 1/(12*x) */
x='x+O('x^55); S152537=eta(x)/(12*x);


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



