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A136118 Least index m>0 such that A136117(n)-A000326(m) is again a pentagonal number. 11

%I #4 May 29 2015 06:29:36

%S 5,4,7,12,19,17,25,20,10,28,45,42,39,17,37,21,36,35,13,33,65,28,67,32,

%T 52,40,74,31,70,85,35,16,60,70,77,68,42,30,105,76,59,26,74,49,115,19,

%U 125,115,102,110,92,56,103,29,145,100,114,77,92,47,63,108,152,95,22,116

%N Least index m>0 such that A136117(n)-A000326(m) is again a pentagonal number.

%e a(1)=5 is the least integer m>0 such that A136117(1)-P(m) is a pentagonal number, namely P(7)-P(5)=70-35=35=P(5).

%e a(2)=4 is the least integer m>0 such that A136117(2)-P(m) is a pentagonal number, namely P(8)-P(4)=92-22=70=P(7).

%o (PARI) A136118vect(n,i=-1)=vector(n,k,until(0,for(j=2,#n=sum2sqr((i+=6)^2+1),n[j]%6==[5,5]||next;n=n[j];break(2)));n[1]\6+1) /* This uses sum2sqr(), cf. A133388. Below some simpler but much slower code. */

%o my(P=A000326(n)=n*(3*n-1)/2,isPent(t)=P(sqrtint(t*2\3)+1)==t); for(i=1,299,for(j=1,(i+1)\sqrt(2),isPent(P(i)-P(j))&print1(j",")||next(2)))

%Y Cf. A000326, A136112-A136117.

%K nonn

%O 1,1

%A _M. F. Hasler_, Dec 25 2007

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