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Search: id:A081458
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| A081458 |
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Sequence of quotients ranging through differing even values of m,n after one division by 2 of numbers of the form 3^m + 5^n, for m,n=0,2,4,6... |
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+0
1
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| 1, 13, 313, 7813, 195313, 4882813, 122070313, 3051757813, 76293945313, 1907348632813, 47683715820313, 17, 317, 7817, 195317, 4882817, 122070317, 3051757817, 76293945317, 1907348632817, 47683715820317, 353, 7853, 195353
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Except for the first term, these numbers always end in 3 and 7 and necessarily generate an odd number as the quotient upon a single division by 2. Indeed for even m,n 3^m+5^n can be written as (4-1)^m + (4+1)^n = 4h+1 + 4i+1 for some h,i. Then we add and get 4(h+i)+2. Divide by 2 to get 2(h+i) + 1 an odd number. We dispose of the endings > 1 being either 3 or 7 by noting that the ending digits of even powers of 3 are 1,9,1,9,... and ending digits of powers of 5 end in 5. Then when we add 1 and 5 we get 6 and 6/2 =3. Similarly, 9+5 = 14. 14/2=7.
The terms are part of a Pythagorean triple: sqrt((a(n))^2 - (a(n)-1)^2)) = 5^n. E.g. sqrt(313^2 - 312^2) = 5^2 since 313 = a(2). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 27 2006
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FORMULA
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a(n) = 26*a(n-1) - 25*a(n-2), n>1. Let M = the 2 X 2 matrix [13, 12; 12, 13]. Then a(n) = left term in M^n *[1,0]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 27 2006
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PROGRAM
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(PARI) p3mp5n(m, n) = { forstep(x=0, m, 2, forstep(x1=0, n, 2, y = (3^x + 5^x1)/2; print1(y" ") ) ) }
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CROSSREFS
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Sequence in context: A009119 A142425 A119131 this_sequence A052114 A132485 A012492
Adjacent sequences: A081455 A081456 A081457 this_sequence A081459 A081460 A081461
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KEYWORD
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easy,nonn
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AUTHOR
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Cino Hilliard (hillcino368(AT)gmail.com), Apr 20 2003
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