A261132 -id:A261132 - cp's OEIS frontend

A262526 Any number greater than a(n) can conjecturally be represented in more ways by sums of three base-10 palindromes than a(n).

1, 2, 3, 4, 98, 120, 142, 164, 172, 192, 212, 223, 2082, 2102, 2203, 2213, 130282, 130992, 131392, 131492, 131592, 131742, 131752, 131792, 131902, 132002, 132102, 132192, 132202, 132482, 132502, 132602, 132662, 132672, 132752, 132782, 132802

Offset: 1

Hugo Pfoertner, Sep 25 2015

The corresponding representation counts are provided in A262527. Positions of latest occurrence of increasing minima of representation counts in A261132. The sequence provides numerical evidence for the validity of the conjecture that every number is the sum of three palindromes.

			a(5)=98 because A261132(k)>5 for all k>98.
a(7)=142 because A261132(k)>A262527(7)=8 for all k>142.

Hugo Pfoertner, Table of n, a(n) for n = 1..1072

Cf. A262527, A261132, A002113, A262524, A262525.

See A261422, A262544, A262545 for another approach.

A262527 Conjectured minimum number of ways to represent a number >= A262526(i) by sums of three base-10 palindromes.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 11, 13, 14, 15, 16, 17, 18, 21, 27, 29, 35, 41, 45, 48, 51, 54, 55, 56, 61, 63, 64, 65, 74, 75, 79, 80, 82, 83, 85, 86, 87, 91, 111, 112, 113, 114, 115, 116, 118, 120, 121, 124, 127, 133, 134, 138, 140, 141, 142, 145, 147, 150, 153, 165, 169, 171, 174, 175, 177, 179, 180, 183, 184, 185

Offset: 1

Hugo Pfoertner, Sep 25 2015

The sequence is obtained by sorting the counts A261132 into increasing order together with their positions of occurrence. If a new record in the sorted A261132 is found, the index of its latest occurrence in A261132 becomes the next term in A262526 and the corresponding value of A261132 becomes a(n).
7 is not in the sequence, because the latest occurrence of 7 is at A261132(64), whereas the latest occurrence of 6 had already set the record to A262526(6) = 120.
a(7) = 8 corresponds to the latest occurrence of 8 at A261132(142), thus A262526(7) = 142.

			a(5) = 5 because A261132(k) > 5 for all k > A262526(5) = 98.

Hugo Pfoertner, Table of n, a(n) for n = 1..1072
Hugo Pfoertner, Least number of representations by sums of three palindromes.

Cf. A002113, A261132, A262524, A262525, A262526.

See A261422, A262544, A262545 for another approach.

A290335 Number of representations of n as a sum of four terms of A020330 (including 0), where order matters.

Original entry on oeis.org

1, 0, 0, 4, 0, 0, 6, 0, 0, 4, 4, 0, 1, 12, 0, 4, 12, 0, 12, 4, 6, 12, 0, 12, 4, 12, 6, 0, 24, 0, 10, 12, 0, 16, 0, 12, 10, 0, 12, 12, 13, 0, 12, 12, 0, 16, 12, 0, 16, 24, 6, 24, 12, 0, 32, 16, 12, 24, 24, 12, 25, 36, 0, 32, 36, 12, 40, 24, 12, 36, 36, 12, 34, 36, 12, 40, 36, 12, 30, 36, 12, 40, 36, 12, 52, 24, 12, 36, 24, 12, 34, 48, 6, 52, 36, 0, 54, 12, 12

Offset: 0

Jeffrey Shallit, Jul 27 2017

			For n = 24 there are four representations, which are the distinct permutations of [15,3,3,3].

Amiram Eldar, Table of n, a(n) for n = 0..10000
Parthasarathy Madhusudan, Dirk Nowotka, Aayush Rajasekaran and Jeffrey Shallit, Lagrange's Theorem for Binary Squares, in: I. Potapov, P. Spirakis and J. Worrell (eds.), 43rd International Symposium on Mathematical Foundations of Computer Science (MFCS 2018), Schloss Dagstuhl, 2018, pp. 18:1-18:14; arXiv preprint, arXiv:1710.04247 [math.NT], 2017-2018.

Cf. A020330, A261132, A290334, A298731.

v = Table[k + k * 2^Floor[Log2[k] + 1], {k, 0, 8}]; a[n_] := If[(ip = IntegerPartitions[n, {4}, v]) == {}, 0, Plus @@ Length /@ (Permutations /@ ip)]; Table[a[n], {n, 0, v[[-1]]}] (* Amiram Eldar, Apr 09 2021 *)

A282585 Number of ways to write n as an ordered sum of 3 squarefree palindromes (A071251).

Original entry on oeis.org

0, 0, 0, 1, 3, 6, 7, 9, 12, 19, 21, 21, 18, 24, 27, 28, 18, 18, 19, 24, 15, 10, 6, 12, 12, 12, 9, 9, 12, 15, 18, 12, 9, 7, 15, 15, 15, 9, 12, 15, 18, 18, 12, 9, 9, 18, 15, 12, 0, 9, 9, 9, 0, 0, 0, 6, 6, 9, 12, 9, 12, 15, 18, 18, 12, 9, 13, 18, 18, 18, 9, 15, 18, 21, 18, 12, 9, 15, 21, 21, 21, 9, 18, 21, 24, 18

Offset: 0

Ilya Gutkovskiy, Feb 19 2017

Every number can be written as the sum of 3 palindromes (see A261132 and A261422).
Conjecture: a(n) > 0 for any sufficiently large n.
Additional conjecture: every number > 3 can be written as the sum of 4 squarefree palindromes.

			a(22) = 6 because we have [11, 6, 5], [11, 5, 6] [6, 11, 5], [6, 5, 11], [5, 11, 6] and [5, 6, 11].

Ilya Gutkovskiy, Extended graphical example
Ilya Gutkovskiy, Extended graphical example for additional conjecture
Eric Weisstein's World of Mathematics, Palindromic Number
Eric Weisstein's World of Mathematics, Squarefree
Index entries for sequences related to palindromes

Cf. A002113, A005117, A035137, A071251, A091580, A091581, A260254, A261131, A261132, A261422, A280210, A282584.

nmax = 85; CoefficientList[Series[Sum[Boole[SquareFreeQ[k] && PalindromeQ[k]] x^k, {k, 1, nmax}]^3, {x, 0, nmax}], x]

G.f.: (Sum_{k>=1} x^A071251(k))^3.

A261916 Smallest p such that n can be written as n = p+q+r where p>=q>=r>=0 are palindromes.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 11, 11, 11, 11, 22, 11, 22, 22, 22, 22, 22, 22, 22, 22, 22, 33, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 44, 22, 33, 33, 33, 33, 33, 33, 33, 33, 33, 55, 22, 33, 33, 33, 33, 33

Offset: 0

N. J. A. Sloane, Sep 11 2015

Every number is the sum of three palindromes.

			Initial values of n,p,q,r are:
0 0 0 0
1 1 0 0
2 1 1 0
3 1 1 1
4 2 1 1
5 2 2 1
6 2 2 2
7 3 3 1
...
25 9 9 7
26 9 9 8
27 9 9 9
28 11 11 6
29 11 11 7
30 11 11 8
...
33 11 11 11
34 22 11 1
...

David Consiglio, Jr., Table of n, a(n) for n = 0..10000
Javier Cilleruelo, Florian Luca and Lewis Baxter, Every positive integer is a sum of three palindromes, arXiv: 1602.06208 [math.NT], 2017, Math. Comp. 87 (2018), 3023-3055.
David Consiglio, Jr., Python program
James Grime and Brady Haran, Every Number is the Sum of Three Palindromes, Numberphile video (2018)

Cf. A002113, A261422, A261132.

If "smallest" is changed to "largest" we get a sequence which agrees with the palindromic floor function A261423 for at least 300 terms.

Edited by Alois P. Heinz, Dec 29 2018

A262528 Maximum number of backward steps k needed to find a representation of an n-digit decimal number x as a sum of three base-10 palindromes of the form k-th largest base-10 palindrome <= x plus a number representable as sum of two base-10 palindromes from A260255.

Original entry on oeis.org

0, 1, 1, 3, 3, 11, 4, 10, 4, 23, 9, 15, 6, 23, 11

Offset: 1

Hugo Pfoertner, Sep 26 2015

The sequence terms are counterexamples to the second part of the claim stated in the answers to the Math Magic Problem of the Month (June 1999) that "all sufficiently large numbers seem to be the sum of 3 palindromes, one of which is the biggest or second biggest possible", which would mean all a(k)=2 for k "sufficiently large".
Since exhaustive search is currently (2015) considered as not feasible, a(16)>=16, a(17)>=7, a(18)>=25, a(19)>=14 are only lower bounds for the next sequence terms.
M. Sigg has shown that a(n)>=3 for n = 5 + 4 * j.

			a(1)=0 because all 1-digit numbers are palindromes,
a(2)=a(3)=1 because all 2-digit and all 3-digit numbers can be represented by the nearest smaller palindrome and a number <=10, e.g., 201=191+9+1.
a(4)=3, because for the number 2023 the largest palindrome leading to a difference representable as sum of two palindromes is 1881. 2023-2002=21 and 2023-1991=32 are not in A260255. 2023-1881=142=141+1 is in A260255. No other 4-digit number requires more than 3 backward steps.
a(6)=11 because for the 6-digit number 101199 none of the first 10 differences 101199-101101=98, 101199-10001=1198, 101199-99999=1200, 101199-99899=1300, 101199-99799=1400, 101199-99699=1500, 101199-99599=1600, 101199-99499=1700, 101199-99399=1800, 101199-99299=1900 is representable as sum of two palindromes (i.e., are in A035137), whereas the 11th palindrome 99199 leads to 101199-99199=2000=1991+9.
a(18)>=25 because for the number x=100000001814566071 only the 25th palindrome < x 99999997779999999 produces the first difference 4034566072 representable as sum of 2 palindromes.

Erich Friedman, Problem of the Month (June 1999)
Markus Sigg, On a conjecture of John Hoffman regarding sums of palindromic numbers, arXiv:1510.07507 [math.NT], 2015.

Cf. A002113, A035137, A088601, A109326, A260255, A261132, A261675.

A282845 Number of ways to write n as an ordered sum of 6 prime power palindromes (A084092).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 6, 21, 56, 126, 246, 432, 702, 1077, 1576, 2232, 3072, 4112, 5352, 6801, 8422, 10197, 12102, 14117, 16146, 18177, 20112, 21882, 23382, 24661, 25566, 26136, 26316, 26181, 25560, 24677, 23436, 21981, 20226, 18486, 16536, 14642, 12702, 10962, 9166, 7662, 6222, 5042, 3912, 3096, 2306, 1746, 1236, 921, 600

Offset: 0

Ilya Gutkovskiy, Feb 22 2017

Is there k which satisfies a(n) > 0 for all n > k?

			a(7) = 6 because we have:
[2, 1, 1, 1, 1, 1]
[1, 2, 1, 1, 1, 1]
[1, 1, 2, 1, 1, 1]
[1, 1, 1, 2, 1, 1]
[1, 1, 1, 1, 2, 1]
[1, 1, 1, 1, 1, 2]

Ilya Gutkovskiy, Extended graphical example
Eric Weisstein's World of Mathematics, Prime Power
Eric Weisstein's World of Mathematics, Palindromic Number
Index entries for sequences related to palindromes

Cf. A000961, A002113, A002385, A084092, A260254, A261131, A261132, A261422, A282585.

nmax = 55; CoefficientList[Series[(x + Sum[Boole[PrimePowerQ[k] && PalindromeQ[k]] x^k, {k, 1, nmax}])^6, {x, 0, nmax}], x]

G.f.: (Sum_{k>=1} x^A084092(k))^6.

A319439 Number of ways to write n as the sum, u + v + w, of three base-2 palindromes (from A006995) with 0 <= u <= v <= w.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 3, 2, 4, 3, 4, 2, 4, 2, 6, 2, 6, 3, 6, 2, 7, 3, 7, 3, 7, 2, 9, 2, 7, 3, 9, 3, 10, 4, 10, 4, 11, 3, 12, 2, 11, 3, 11, 1, 12, 2, 10, 4, 11, 2, 14, 3, 11, 4, 13, 1, 13, 2, 11, 3, 12, 2, 15, 2, 13, 5, 14, 3, 17, 2, 13, 4, 15, 2, 17, 2, 12, 4, 15

Offset: 0

Peter Kagey, Sep 18 2018

Every number n can be written as the sum of four base-2 palindromes.

a(A261678(n)) = 0.

			a(13) = 4 because 13 can be written as the sum of three base-2 palindromes in four different ways:
13 = 5 + 5 + 3 = 101_2  + 101_2 + 11_2,
13 = 7 + 3 + 3 = 111_2  +  11_2 + 11_2,
13 = 7 + 5 + 1 = 111_2  + 101_2 +  1_2, and
13 = 9 + 3 + 1 = 1001_2 +  11_2 +  1_2.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
A. Rajasekaran, J. Shallit, T. Smith, Sums of Palindromes: an Approach via Automata, arXiv:1706.10206 [cs.FL], 2017.
Rémy Sigrist, PARI program for A319439

Cf. A006995, A261132, A261678.

PARI
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See Links section.
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a(0) corrected by Rémy Sigrist, Sep 19 2018

cp's OEIS Frontend

A262526 Any number greater than a(n) can conjecturally be represented in more ways by sums of three base-10 palindromes than a(n).

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A262527 Conjectured minimum number of ways to represent a number >= A262526(i) by sums of three base-10 palindromes.

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A290335 Number of representations of n as a sum of four terms of A020330 (including 0), where order matters.

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A282585 Number of ways to write n as an ordered sum of 3 squarefree palindromes (A071251).

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A261916 Smallest p such that n can be written as n = p+q+r where p>=q>=r>=0 are palindromes.

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A262528 Maximum number of backward steps k needed to find a representation of an n-digit decimal number x as a sum of three base-10 palindromes of the form k-th largest base-10 palindrome <= x plus a number representable as sum of two base-10 palindromes from A260255.

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A282845 Number of ways to write n as an ordered sum of 6 prime power palindromes (A084092).

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A319439 Number of ways to write n as the sum, u + v + w, of three base-2 palindromes (from A006995) with 0 <= u <= v <= w.

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