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Antidiagonal sums of square table A112564 of generalized Flavius Josephus sieves.
6

%I #17 Jun 23 2020 11:03:16

%S 1,2,4,9,19,43,88,207,423,951,2094,4511,9445,22025,45172,93483,205101,

%T 435515,882322,1983637,3983429,8659805,17928712,36742421,76317521,

%U 163095069,331056988,693360803,1444266731

%N Antidiagonal sums of square table A112564 of generalized Flavius Josephus sieves.

%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>

%o (C++) #include <iostream> #include <limits.h> using namespace std ; int A112564(const unsigned long long n, const unsigned long long k) { unsigned long long A=k ; unsigned long long B=0 ; unsigned long long C=0 ; if( n==0 || k == 0) return 1; else { while(A != B) { C++ ; if ( C % n == 0) C++ ; B=A ; A= A*(C+1)/C ; if ( A > ULLONG_MAX/(C+1) ) exit(0) ; } return 1+A ; } } unsigned long long A112569(const int d) { unsigned long long resul =0 ; for(int k=0 ; k <=d ; k++) resul += A112564(d-k,k) ; return resul ; } int main(int argc, char *argv[]) { for(int n=0; n < 40 ; n++) cout << A112569(n) << endl ; } - _R. J. Mathar_, Sep 26 2006

%o (PARI) A112564(n, k)= {my(A=k, B=0, C=0); if(n==0 || k==0, 1, until(A==B, C=C+1; if(C%n==0, C=C+1); B=A; A=floor(A*(C+1)/C)); 1+A)}; a(n) = sum(k=0, n, A112564(n-k, k)); \\ _Michel Marcus_, Apr 23 2013

%Y Cf. A000960, A112564.

%K more,nonn

%O 1,2

%A _Paul D. Hanna_, Oct 14 2005

%E More terms from _R. J. Mathar_, Sep 26 2006