A262503 a(n) = largest k such that A155043(k) = n.
0, 2, 6, 12, 18, 22, 30, 34, 42, 48, 60, 72, 84, 96, 108, 120, 132, 140, 112, 116, 126, 124, 130, 138, 150, 156, 168, 180, 176, 184, 192, 204, 216, 228, 240, 248, 264, 280, 250, 258, 270, 288, 296, 312, 306, 320, 328, 340, 352, 364, 372, 354, 358, 368, 384, 396, 420, 402, 414, 418, 432, 450, 468, 480, 504, 520, 540, 560, 572, 580, 594, 612, 610, 618, 622, 628, 648, 672, 592
Offset: 0
Keywords
Links
- Antti Karttunen, Table of n, a(n) for n = 0..110880
- A. Karttunen, Ratio A262502(n+2)/a(n) drawn with the help of OEIS Plot2-script
- A. Karttunen, Ratio a(n)/A261089(n) drawn with the help of OEIS Plot2-script
Crossrefs
Programs
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Mathematica
lim = 80; a[0] = 0; a[n_] := a[n] = 1 + a[n - DivisorSigma[0, n]]; t = Table[a@ n, {n, 0, 12 lim}]; Last@ Flatten@ Position[t, #] - 1 & /@ Range[0, lim] (* Uses the product of a limit and an arbitrary coefficient (12) based on observation of output for low values (n < 500). This might need to be adjusted for large n to give correct values of a(n). - Michael De Vlieger, Sep 29 2015 *) (* Note: one really should use a general safe limit, like A262502(n+2) I use in my Scheme-program. - Antti Karttunen, Sep 29 2015 *)
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PARI
allocatemem(123456789); uplim = 2162160; \\ = A002182(41). v155043 = vector(uplim); v155043[1] = 1; v155043[2] = 1; for(i=3, uplim, v155043[i] = 1 + v155043[i-numdiv(i)]); A155043 = n -> if(!n,n,v155043[n]); uplim2 = 110880; \\ = A002182(30). v262503 = vector(uplim2); for(i=1, uplim, if(v155043[i] <= uplim2, v262503[v155043[i]] = i)); A262503 = n -> if(!n,n,v262503[n]); for(n=0, uplim2, write("b262503.txt", n, " ", A262503(n)));
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Scheme
(define (A262503 n) (let loop ((k (A262502 (+ 2 n)))) (if (= (A155043 k) n) k (loop (- k 1)))))
Formula
Other identities and observations. For all n >= 0:
A262502(n+2) > a(n). [Not rigorously proved, but empirical evidence and common sense agrees.]
Comments