cp's OEIS Frontend

A144954 a(n) = start of a sequence of at least n consecutive primes, p_1, p_2, ..., p_n (say), all == 1 mod 4, such that A(p_1) > A(p_2) > ... > A(p_n), where A(p) (see A145010) is the area of the Pythagorean triangle with hypotenuse p.

%I A144954 #13 Sep 10 2018 14:38:39
%S A144954 5,37,157,1277,4441,8669,14533,883241,10006957,530551397,931953301,
%T A144954 931953301
%N A144954 a(n) = start of a sequence of at least n consecutive primes, p_1, p_2, ..., p_n (say), all == 1 mod 4, such that A(p_1) > A(p_2) > ... > A(p_n), where A(p) (see A145010) is the area of the Pythagorean triangle with hypotenuse p.
%C A144954 Prompted by a question from Shiv K. Gupta to the Number Theory mailing list.
%H A144954 D. Broadhurst in reply to S. K. Gupta, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;c980747e.0902">Hypotenuses of Integral Right Triangles - again</a>, NMBRTHRY list, Feb 24 2009
%F A144954 a(n) = min { A002144(k) | A145010(k) > A145010(k+1) > ... > A145010(k+n-1)}. - _M. F. Hasler_, Feb 26 2009
%e A144954 Comment from _M. F. Hasler_, Feb 24 2009:
%e A144954 The first sequence of 12 such primes is the one starting at a(12) =
%e A144954 931953301 = [27050, 14151]^2 ; area = 203431499448450450
%e A144954 931953389 = [26050, 15917]^2 ; area = 176325413694076350
%e A144954 931953397 = [25239, 17174]^2 ; area = 148267841956285170
%e A144954 931953409 = [24528, 18175]^2 ; area = 120941067830427600
%e A144954 931953433 = [30332, 3453 ]^2 ; area = 95111855933417940
%e A144954 931953437 = [23846, 19061]^2 ; area = 93319265825216970
%e A144954 931953469 = [30462, 2005 ]^2 ; area = 56429222392003890
%e A144954 931953509 = [30478, 1745 ]^2 ; area = 49241224048436490
%e A144954 931953569 = [30487, 1580 ]^2 ; area = 44651199683914740
%e A144954 931953637 = [22166, 20991]^2 ; area = 23594434443844350
%e A144954 931953709 = [30525 , 422 ]^2 ; area = 12000420304268550
%e A144954 931953733 = [21793, 21378]^2 ; area = 8346882442487610
%o A144954 (PARI) A144954( n, p=5, verbose=0, L=[0])={ for( i=1,n-1, while(( p=nextprime(p+2)) % 4 !=1,); mn=sum2sqr_prime(p); L=if( L[i] > A=mn[1]*mn[2]*abs(mn[1]^2-mn[2]^2), concat( L, A), i=0; [A]) ); for( i=0,n-1, i & while( 1 != (p=precprime(p-2)) % 4,); verbose & print( p" = " sum2sqr_prime(p) "^2 ; area = " L[n-i])); p} \\ _M. F. Hasler_, Feb 24 2009
%Y A144954 Cf. A145010, A002144, A002330, A002331. See A144960 for the actual primes.
%K A144954 nonn
%O A144954 1,1
%A A144954 _David Broadhurst_, Feb 24 2009