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A144954
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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.
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4
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5, 37, 157, 1277, 4441, 8669, 14533, 883241, 10006957, 530551397, 931953301, 931953301
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OFFSET
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1,1
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COMMENTS
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Prompted by a question from Shiv K. Gupta to the Number Theory mailing list.
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LINKS
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FORMULA
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EXAMPLE
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The first sequence of 12 such primes is the one starting at a(12) =
931953301 = [27050, 14151]^2 ; area = 203431499448450450
931953389 = [26050, 15917]^2 ; area = 176325413694076350
931953397 = [25239, 17174]^2 ; area = 148267841956285170
931953409 = [24528, 18175]^2 ; area = 120941067830427600
931953433 = [30332, 3453 ]^2 ; area = 95111855933417940
931953437 = [23846, 19061]^2 ; area = 93319265825216970
931953469 = [30462, 2005 ]^2 ; area = 56429222392003890
931953509 = [30478, 1745 ]^2 ; area = 49241224048436490
931953569 = [30487, 1580 ]^2 ; area = 44651199683914740
931953637 = [22166, 20991]^2 ; area = 23594434443844350
931953709 = [30525 , 422 ]^2 ; area = 12000420304268550
931953733 = [21793, 21378]^2 ; area = 8346882442487610
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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