cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A248134 Consider a number x as a concatenation of two integers, a and b: x = concat(a,b). Take their sum and repeat the process deleting the minimum number and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.

Original entry on oeis.org

14, 19, 21, 28, 42, 47, 63, 84, 105, 126, 147, 149, 168, 189, 199, 298, 323, 497, 646, 795, 911, 969, 1292, 1499, 1822, 1999, 2087, 2733, 2998, 3089, 3248, 3379, 3644, 4555, 4997, 5411, 5466, 6178, 6377, 6496, 7288, 7995, 8199, 9161, 9267, 9744, 10822, 12356
Offset: 1

Views

Author

Paolo P. Lava, Oct 02 2014

Keywords

Comments

If the number x is rewritten as concat(a,b), the problem is to find a value of y such that x = a*F(y) + b*F(y+1), if a < b, or x = b*F(y) + a*F(y+1), if a > b, where F(y) is a Fibonacci number (see values of x, a, b, y, for 1
Similar to A130792 but here the minimum number is deleted since the beginning.
All the listed numbers admit only one concatenation, concat(a,b), that, through the addition process, leads to themselves. Is there any number that admit more than one single concatenation?
Sequence is infinite. Let us consider the numbers 19, 199, 1999, 19...9 and let us divide them as concat(1,9), concat(1,99), concat(1,999), concat(1,9...9). In two steps we have the initial numbers back: 1 + 9 = 10 and 9 + 10 = 19; 1 + 99 = 100 and 99 + 100 = 199, etc.

Examples

			Let us rewrite 5411 as 54 U 11. Then:
11 + 54 = 65;
54 + 65 =  119;
65 + 119 = 184;
119 + 184 = 303;
184 + 303 = 487;
303 + 487 = 790;
487 + 790 = 1277;
790 + 1277 = 2067;
1277 + 2067 = 3344;
2067 + 3344 = 5411, that is 11*F(10) + 54*F(11) = 11*55 + 54*89 = 605 + 4806 = 5411.

Links

Crossrefs

Cf. A000045, A007629, A130792, A246544, A247012, A247013.

Programs

  • Maple
    P:=proc(q,h) local a,b,k,n,t,v; v:=array(1..h);
for n from 1 to q do for k from 1 to ilog10(n) do
a:=n mod 10^k; b:=trunc(n/10^k); if a

A307863 Numbers x = concat(a,b) such that b and a are the first two terms for a Fibonacci-like sequence containing x itself.

Original entry on oeis.org

17, 21, 25, 42, 63, 84, 105, 123, 126, 147, 168, 189, 197, 246, 295, 369, 492, 787, 1033, 1115, 1141, 1248, 1279, 1997, 2066, 2230, 2282, 2496, 2995, 3099, 3345, 3423, 3744, 4460, 4564, 4992, 5411, 5575, 5705, 6690, 6846, 7987, 10112, 10483, 10822, 11059, 11107
Offset: 1

Views

Author

Paolo P. Lava, May 02 2019

Keywords

Comments

Similar to A130792 but here the sums start b + a = c, a + c = d, etc.
First six terms are also the first six Inrepfigit numbers (A128546).
Being x = concat(a,b), the problem is to find an index y such that x = b*F(y) + a*F(y+1), where F(y) is a Fibonacci number (see file with values of x, b, a, y, for 1< x <10^6, in Links). All the listed numbers admit only one unique concatenation that, through the addition process, leads to themselves. Is there any number that admits more than one single concatenation?

Examples

			123 can be split into 1 and 23 and the Fibonacci-like sequence: 23, 1, 24, 25, 49, 74, 123, ... contains 123 itself.

Links

Crossrefs

Cf. A128546, A130792, A289868.

Programs

  • Maple
    P:=proc(n) local j,t,v; v:=array(1..100);
for j from 1 to length(n)-1 do v[1]:=n mod 10^j; v[2]:=trunc(n/10^j);
v[3]:=v[1]+v[2]; t:=3; while v[t]Paolo P. Lava, May 02 2019

A383230 Numbers k whose decimal representation can be split in three parts which can be used as seeds for a tribonacci-like sequence containing k itself.

Original entry on oeis.org

197, 742, 1007, 1257, 1484, 1749, 1789, 3241, 4349, 4515, 4851, 5709, 6482, 6925, 7756, 8196, 8449, 8698, 10232, 10997, 11627, 16898, 17206, 18353, 19789, 20464, 27315, 30696, 31385, 35537, 40928, 43367, 44111, 48310, 48591, 49228, 50574, 58506, 62770, 79976, 88222
Offset: 1

Views

Author

Paolo P. Lava, Apr 20 2025

Keywords

Comments

There are 306 terms < 10^8.
If the number k is rewritten as the concatenation of a, b and c, the problem is to find an integer x such that k = a*A000073(x) + b*A001590(x+1) + c*A000073(x+1).
Is there any term that results from more than one concatenation?

Examples

			1007 can be split into 10, 0, 7 and the tribonacci-like sequence contains 1007 itself: 10, 0, 7, 17, 24, 48, 89, 161, 298, 548, 1007 ... (x = 9, as per second comment);
1257 can be split into 1, 25, 7 and the tribonacci-like sequence contains 1257 itself: 1, 25, 7, 33, 65, 105, 203, 373, 681, 1257 ... (x = 8, as per second comment);
16898 can be split into 16, 8, 98 and the tribonacci-like sequence contains 16898 itself: 16, 8, 98, 122, 228, 448, 798, 1474, 2720, 4992, 9186, 16898 ... (x = 10, as per second comment).

Links

Crossrefs

Cf. A000073, A000213, A001590, A007629, A130792.

Programs

  • Maple
    P:=proc(q) local b,c,d,f1,f2,f3,i,j,m,n,t,v,y,x,w; i:=[]; for n from 100 to q do b:=length(n);
for t from 1 to b-2 do c:=n mod 10^t; m:=trunc(n/10^t); d:=length(m);
for j from 1 to d-1 do x:=trunc(m/10^j); y:=m mod 10^j; f1:=2; f2:=3; f3:=4; v:=x*f1+y*f2+c*f3;
while v
Showing 1-3 of 3 results.

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