A249401 - OEIS

A249401

a(n) = OP(sum{i=0,...,n} OP(binomial(n,i))), where OP(n) is the odd part of n (A000265).

1, 1, 3, 1, 7, 11, 43, 1, 67, 103, 405, 157, 1603, 2627, 10819, 1, 11427, 18395, 73273, 27099, 245929, 385333, 1516069, 8955, 3288479, 5588411, 24411113, 9596521, 104620147, 171729011, 708599923, 1, 496472227, 846762835, 3610797801, 1364604931, 13485280597

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

In 1899, Glaisher, in particular, proved that the number of odd entries in the n-th row of Pascal triangle (A007318) is 2^s(n) (A001316), where s(n) is the number of 1's in the binary representation of n (s=A000120). Hence, all entries in the n-th row of Pascal triangle are odd if and only if n has the form 2^k-1, k=0,1,2,..., i.e., n is in A000225. In this case a(n) = OP(2^n) = 1. Thus the sequence contains infinitely many 1's.

REFERENCES

J. Glaisher, On the residue of a binomial-theorem coefficient with respect to a prime modulus, Quarterly J. of Pure and Applied Math. 30 (1899), 150-156.

LINKS

Table of n, a(n) for n=0..36.

FORMULA

a(n) < 2^n.

EXAMPLE

{binomial(4,i)} = {1,4,6,4,1} -> {1,1,3,1,1}. Thus a(4)=1+1+3+1+1=7.

MATHEMATICA

oddPart[n_]:=n/2^IntegerExponent[n, 2];

Table[oddPart[Sum[oddPart[Binomial[n, i]], {i, 0, n}]], {n, 0, 50}] (* Peter J. C. Moses, Oct 27 2014 *)

PROG

(PARI) op(n) = n / 2^valuation(n, 2);

a(n) = op(sum(i=0, n, op(binomial(n, i)))); \\ Michel Marcus, Oct 31 2014