A330128 a(n) is the number of terms in the analog of A121805 but starting with n, or -1 if that sequence is infinite.

2137453, 194697747222394, 2, 199900, 19706, 209534289952018960, 15, 198104936410, 19511030, 20573, 20572, 2137452, 20534, 19238, 2, 2089707, 20670629294, 1, 21482, 19278442756937613, 2074, 19278442756937612, 20571, 194697747222393, 193, 197062, 1, 197, 2061823

Offset: 1

Author

Giovanni Resta, Dec 02 2019

Keywords

• nonn
• base

Comments

The final terms of the corresponding sequences are given in A330129.

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000
• Eric Angelini, Michael S. Branicky, Giovanni Resta, N. J. A. Sloane, and David W. Wilson, The Comma Sequence: A Simple Sequence With Bizarre Properties, arXiv:2401.14346, Fibonacci Quarterly 62:3 (2024), 215-232.
• Eric Angelini, Michael S. Branicky, Giovanni Resta, N. J. A. Sloane, and David W. Wilson, The Comma Sequence: A Simple Sequence With Bizarre Properties, Local copy.
• N. J. A. Sloane, Eric Angelini's Comma Sequence, Experimental Math Seminar, Rutgers Univ., January 18, 2024, Youtube video; Slides

Crossrefs

Cf. A330129 (corresponding last term), A121805, A139284, A366492.

For record high points see A367364 and A367365.

Programs

Mathematica

nxt[x_] := Block[{p=1, n=x}, While[n >= 10, n = Floor[n/10]; p *= 10]; p (n + 1)]; a[n_] := Block[{nT=1, nX=n, w1, w2, w3, x, it, stp, oX}, stp = 100; w1 = w2 = w3 = 0; While[True, oX = nX; nT++; x = 10*Mod[oX, 10]; nX = SelectFirst[Range[9], IntegerDigits[oX + x + #][[1]] == # &, 0]; If[nX == 0, Break[], nX = nX + oX + x]; If[nT == stp, stp += 100; w1=w2; w2=w3; w3=nX; If[w3 + w1 == 2 w2 && Mod[w3 - w2, 100] == 0, it = Floor[(nxt[nX] - nX - 1)/(w3 - w2)]; nT += it*100; nX += (w3 - w2)*it; w3=nX; stp += it*100]]]; nT - 1]; Array[a, 30]

Python

def nxt(x):
    p, n = 1, x
    while n >= 10:
        n //= 10
        p *= 10
    return p * (n + 1)
def a(n):
    nT, nX, w1, w2, w3, stp  = 1, n, 0, 0, 0, 100
    while True:
        oX = nX
        nT += 1
        x = 10*(oX%10)
        nX = next((y for y in range(1, 10) if str(oX+x+y)[0] == str(y)), 0)
        if nX == 0: break
        else: nX += oX + x
        if nT == stp:
            stp += 100
            w1, w2, w3 = w2, w3, nX
            if w3 + w1 == 2*w2 and (w3 - w2)%100 == 0:
                it = (nxt(nX) - nX - 1)//(w3 - w2)
                nT += it*100
                nX += (w3 - w2)*it
                w3 = nX
                stp += it*100
    return nT - 1
print([a(n) for n in range(1, 30)]) # Michael S. Branicky, Nov 18 2023 after Giovanni Resta

Extensions

Escape clause added to definition by N. J. A. Sloane, Nov 14 2023