A261230 -id:A261230 - cp's OEIS frontend

A261234 a(n) = number of steps to reach (3^n)-1 when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

1, 2, 5, 12, 29, 74, 196, 530, 1445, 3956, 10862, 29901, 82592, 229233, 639967, 1797288, 5073707, 14381347, 40890492, 116559600, 333043360, 953890490, 2738788806, 7881915828, 22729464587, 65652788211, 189866467219, 549596773550, 1592118137130, 4615680732717, 13392399641613, 38894563977633, 113074467549440, 329080350818600, 958725278344368, 2795854777347489

Offset: 0

Antti Karttunen, Aug 13 2015

Hiroaki Yamanouchi, Table of n, a(n) for n = 0..100
Antti Karttunen, Naive C-program for computing the terms of A261234, A261236 and A261237 at the same time
Hiroaki Yamanouchi, Fast Python-program for computing terms of A213709, A261234 and analogous sequences in other bases

Cf. A000244, A054861, A261231, A261230.
First differences of A261232 and A261233.
Sum of A261236 and A261237.
Cf. A261235 (first differences of this sequence).
Cf. also A213709.

Table[Length@ NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &] - 1, {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016 *)

a(n) = A261236(n) + A261237(n).

a(23)-a(35) from Hiroaki Yamanouchi, Aug 16 2015

A260215 Expansion of chi(-q) * chi(q^9) / (chi(q) * chi(-q^9)) in powers of q where chi() is a Ramanujan theta function.

Original entry on oeis.org

1, -2, 2, -4, 6, -8, 12, -16, 22, -28, 36, -48, 60, -76, 96, -120, 150, -184, 228, -280, 340, -416, 504, -608, 732, -878, 1052, -1252, 1488, -1768, 2088, -2464, 2902, -3408, 3996, -4672, 5460, -6364, 7400, -8600, 9972, -11544, 13344, -15400, 17752, -20424

Offset: 0

Michael Somos, Aug 13 2015

sign

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

			G.f. = 1 - 2*x + 2*x^2 - 4*x^3 + 6*x^4 - 8*x^5 + 12*x^6 - 16*x^7 + 22*x^8 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..2000
Michael Somos, Introduction to Ramanujan theta functions
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions

Cf. A128143, A260057, A261154, A261156, A261203, A261230, A261325.

a[ n_] := SeriesCoefficient[ QPochhammer[ q, q^2] QPochhammer[ q, -q] QPochhammer[ -q^9, q^18] QPochhammer[ -q^9, q^9], {q, 0, n}];

{a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A)^2 * eta(x^4 + A) * eta(x^18 + A)^3 / (eta(x^2 + A)^3 * eta(x^9 + A)^2 * eta(x^36 + A)), n))};

Expansion of psi(-q) * psi(q^9) / (psi(q) * psi(-q^9)) in powers of q where psi() is a Ramanujan theta function.
Expansion of eta(q)^2 * eta(q^4) * eta(q^18)^3 / (eta(q^2)^3 * eta(q^9)^2 * eta(q^36)) in powers of q.
Euler transform of period 36 sequence [ -2, 1, -2, 0, -2, 1, -2, 0, 0, 1, -2, 0, -2, 1, -2, 0, -2, 0, -2, 0, -2, 1, -2, 0, -2, 1, 0, 0, -2, 1, -2, 0, -2, 1, -2, 0, ...].
G.f. is a period 1 Fourier series which satisfies f(-1 / (36 t)) = g(t) where q = exp(2 Pi i t) and g() is the g.f. of A128143.
a(n) = (-1)^n * A261156(n). Convolution inverse of A261156
a(2*n + 1) = -2 * A261203(n) = -2 * A261154(2*n + 1). 2 * a(2*n) = A261154(2*n) unless n=0.
a(3*n) = A261320(n). a(3*n + 1) = -2 * A261325(n). a(3*n + 2) = 2 * A260057(n). - Michael Somos, Nov 08 2015
a(n) ~ (-1)^n * exp(2*Pi*sqrt(n)/3) / (2*sqrt(3)*n^(3/4)). - Vaclav Kotesovec, Nov 16 2017

A261236 a(n) = A261234(n) - A261237(n).

Original entry on oeis.org

0, 1, 3, 7, 16, 40, 104, 279, 758, 2071, 5678, 15609, 43035, 119139, 331616, 928572, 2614743, 7396880, 20999683, 59784414, 170615755, 488073987, 1399625614, 4023315793, 11590737827, 33452982391

Offset: 0

Antti Karttunen, Aug 16 2015

a(n) = How many numbers whose base-3 representation begins with digit "1" are encountered before (3^n)-1 is reached when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k by k - (sum of digits in base-3 representation of k).

			For n=2, we start from 3^(2+1) - 1 = 26 ("222" in base-3), and subtract 6 to get 20 ("202" in base-3), from which we subtract 4, to get 16 ("121" in base-3), from which we subtract 4, to get 12 ("110" in base-3), from which we subtract 2 to get 10 ("101" in base-3), from which we subtract 2 to get 8 ("22" in base-3), which is the end point of iteration. Of the numbers encountered, 16, 12 and 10 have base-3 representations beginning with digit "1", thus a(2) = 3.

Cf. A261234, A261237, A261230.

/* Use the C-program given in A261234. */

Table[Length@ # - First@ FirstPosition[#, k_ /; k != 2] &@ Map[First@ IntegerDigits[#, 3] &, #] &@ NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &], {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)

(define (A261236 n) (- (A261234 n) (A261237 n)))

a(n) = A261234(n) - A261237(n).

Terms a(24) & a(25) from Antti Karttunen, Jun 27 2016

A261237 Number of steps needed when starting from (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k) to encounter the first number whose base-3 representation begins with a digit other than 2.

Original entry on oeis.org

1, 1, 2, 5, 13, 34, 92, 251, 687, 1885, 5184, 14292, 39557, 110094, 308351, 868716, 2458964, 6984467, 19890809, 56775186, 162427605, 465816503, 1339163192, 3858600035, 11138726760, 32199805820

Offset: 0

Antti Karttunen, Aug 16 2015

a(n) = How many numbers whose base-3 representation begins with digit "2" are encountered before (3^n)-1 is reached when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k by k - (sum of digits in base-3 representation of k). Note that (3^n)-1 (in base-3: "222...", with digit "2" repeated n times) is not included in the count, although the starting point (3^(n+1))-1 is.

			For n=0, we start from 3^(0+1) - 1 = 2 (also "2" in base-3), and subtract 2 to get 0, which doesn't begin with 2, thus a(0) = 1.
For n=1, we start from 3^(1+1) - 1 = 8 ("22" in base-3), and subtract 2*2 = 4 to get 4 ("11" in base-3) which doesn't begin with 2, thus a(1) = 1.
For n=2, we start from 3^(2+1) - 1 = 26 ("222" in base-3), and subtract first 6 to get 20 ("202" in base-3), from which we subtract 4, to get 16 ("121" in base-3), so in two steps we have reached the first such number that does not begin with "2" in base-3, thus a(2) = 2.

Cf. A000244, A054861, A122586, A261230, A261234, A261236.

/* Use the C-program given in A261234. */

Flatten@ Table[FirstPosition[#, k_ /; k != 2] &@ Map[First@ IntegerDigits[#, 3] &, NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &]] - 1, {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)

a(n)=my(k=3^(n+1)-1,t=2*3^n,s); while(k>=t, k-=sumdigits(k,3); s++); s \\ Charles R Greathouse IV, Aug 21 2015

(definec (A261237 n) (let loop ((k (- (A000244 (+ 1 n)) 1)) (s 0)) (if (< (A122586 k) 2) s (loop (* 2 (A054861 k)) (+ 1 s)))))

Terms a(24) & a(25) from Antti Karttunen, Jun 27 2016

cp's OEIS Frontend

A261234 a(n) = number of steps to reach (3^n)-1 when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

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A260215 Expansion of chi(-q) * chi(q^9) / (chi(q) * chi(-q^9)) in powers of q where chi() is a Ramanujan theta function.

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A261236 a(n) = A261234(n) - A261237(n).

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A261237 Number of steps needed when starting from (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k) to encounter the first number whose base-3 representation begins with a digit other than 2.

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C

Mathematica

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