A279543
a(n) = a(n-1) + 3^n * a(n-2) with a(0) = 1 and a(1) = 1.
Original entry on oeis.org
1, 1, 10, 37, 847, 9838, 627301, 22143007, 4137864868, 439978671649, 244776761262181, 78185678507867584, 130162592460442600405, 124783388108159412726037, 622688428086038843429228482, 1791127919536971393223950620041
Offset: 0
1/1 = a(0)/A015460(2).
1/(1+3/1) = 1/4 = a(1)/A015460(3).
1/(1+3/(1+3^2/1)) = 10/13 = a(2)/A015460(4).
1/(1+3/(1+3^2/(1+3^3/1))) = 37/121 = a(3)/A015460(5).
Cf. similar sequences with the recurrence a(n-1) + q^n * a(n-2) for n>1, a(0)=1 and a(1)=1:
A280294 (q=2), this sequence (q=3),
A280340 (q=10).
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RecurrenceTable[{a[n] == a[n - 1] + 3^n*a[n - 2], a[0] == 1, a[1] == 1}, a, {n, 15}] (* Michael De Vlieger, Dec 31 2016 *)
A280294
a(n) = a(n-1) + 2^n * a(n-2) with a(0) = 1 and a(1) = 1.
Original entry on oeis.org
1, 1, 5, 13, 93, 509, 6461, 71613, 1725629, 38391485, 1805435581, 80431196861, 7475495336637, 666367860021949, 123144883455482557, 21958686920654707389, 8092381769059159562941, 2886261393833112966453949, 2124255587862077437434059453
Offset: 0
1/1 = a(0)/A015459(2).
1/(1+2/1) = 1/3 = a(1)/A015459(3).
1/(1+2/(1+2^2/1)) = 5/7 = a(2)/A015459(4).
1/(1+2/(1+2^2/(1+2^3/1))) = 13/31 = a(3)/A015459(5).
Cf. similar sequences with the recurrence a(n-1) + q^n * a(n-2) for n>1, a(0)=1 and a(1)=1: this sequence (q=2),
A279543 (q=3),
A280340 (q=10).
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nxt[{n_,a_,b_}]:={n+1,b,b+2^(n+1)*a}; NestList[nxt,{1,1,1},20][[All,2]] (* Harvey P. Dale, Jul 17 2020 *)
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def a():
a, b, p = 1, 0, 1
while True:
p, a, b = p + p, b, b + p * a
yield b
A280294 = a()
print([next(A280294) for in range(19)]) # _Peter Luschny, Dec 05 2017
Showing 1-2 of 2 results.
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