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A167709
Numbers y such that 19*y^2 + 81 is a square.
12
0, 1, 15, 24, 220, 351, 560, 5124, 8175, 74801, 119340, 190399, 1742145, 2779476, 25432120, 40575249, 64735100, 592324176, 945013665, 8646845999, 13795465320, 22009743601, 201388477695, 321301866624, 2939902207540, 4690417633551
OFFSET
0,3
REFERENCES
A. H. Beiler, "Recreations in the theory of numbers": Ex. 38, page 298 (Dover Publications, Inc., New York, 1966).
FORMULA
G.f.: (z + 15*z^2 + 24*z^3 + 220*z^4 + 351*z^5 + 560*z^6 + 5124*z^7 + 8175*z^8 + 74801*z^9 - 340*z^5*(z + 15*z^2 + 24*z^3 + 220*z^4) ) / (1 - 340*z^5 + z^10).
a(n+10) = 340*a(n+5) - a(n).
a(n+5) = 170*a(n) + 39*sqrt(19*a(n)^2 + 81).
For n == 0 (mod 5): a(n) = ( 9*sqrt(19) )/38*(170 + 39*sqrt(19))^(n) + (-9*sqrt(19))/38*(170 - 39*sqrt(19))^(n); the subsequence is 0, 351, 119340, 40575249, 13795465320, 4690417633551, 1594728199942020, 542202897562653249, 184347390443102162640, ...
For n == 1 (mod 5): a(n) = (10*sqrt(19) + 19)/38*(170 + 39*sqrt(19))^(n) + (-10*sqrt(19) + 19)/38*(170 - 39*sqrt(19))^(n); the subsequence is 1, 560, 190399, 64735100, 22009743601, 7483248089240, 2544282340597999, 865048512555230420, 294113949986437744801, ...
For n == 2 (mod 5): a(n) = (66*sqrt(19) + 285)/38*(170 + 39*sqrt(19))^(n) + (-66*sqrt(19) + 285)/38*(170 - 39*sqrt(19))^(n); the subsequence is 15, 5124, 1742145, 592324176, 201388477695, 68471490092124, 23280105242844465, 7915167311077025976, 2691133605660945987375, ...
For n == 3 (mod 5): a(n) = (105*sqrt(19) + 456)/38*(170 + 39*sqrt(19))^(n) + (-105*sqrt(19) + 456)/38*(170 - 39*sqrt(19))^(n); the subsequence is 24, 8175, 2779476, 945013665, 321301866624, 109241689638495, 37141853175221676, 12628120837885731345, 4293523943027973435624, ...
For n == 4 (mod 5): a(n) = (959*sqrt(19) + 4180)/38*(170 + 39*sqrt(19))^(n) + (-959*sqrt(19) + 4180)/38*(170 - 39*sqrt(19))^(n); the subsequence is 220, 74801, 25432120, 8646845999, 2939902207540, 999558103717601, 339846815361776800, 115546917664900394399, 39285612159250772318860, ...
EXAMPLE
a(0) = 0 because 19*0 + 81 = 9^2, a(1)=1 because 19*1 + 81 = 10^2.
MAPLE
a(0):=0:a(1):=1:a(2):=15:a(3):=24:a(4):=220:a(5):=351:a(6):=560: a(7):=5124: a(8):=8175:a(9):=74801:for n from 0 to 40 do a(n+10):=340*a(n+5)-a(n):od:seq(a(n), n=0..40);
MATHEMATICA
a[0]=0; a[1]=a[-1]=1; a[2]=a[-2]=15; a[n_] := a[n] = 170*a[n-5]+39*Sqrt[19*a[n-5]^2+81]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Dec 19 2013 *)
LinearRecurrence[{0, 0, 0, 0, 340, 0, 0, 0, 0, -1}, {0, 1, 15, 24,
220, 351, 560, 5124, 8175, 74801}, 100] (* G. C. Greubel, Jun 20 2016 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Richard Choulet, Nov 10 2009
STATUS
approved