Matej Veselovac - cp's OEIS frontend

A337671 Subsequence of A337670 in which there are at most five terms in the sum.

1422, 1464, 1554, 2612, 3127, 4481, 5644, 16122, 68521, 77129, 82583, 1065585, 4227140, 6164560

Offset: 1

Matej Veselovac, Sep 28 2020

Number m is in the sequence if there exists a set of unordered {base, exponent} pairs {{b_1, e_1}, ..., {b_k, e_k}}, k <= 5, representing non-commutative perfect powers b_i^e_i != e_i^b_i, b_i > 1, e_i > 1, whose sum equals m = Sum_{i=1..k} b_i^e_i = Sum_{i=1..k} e_i^b_i.
If it exists, what is the smallest term whose sum consists of exactly 2, 3 or 4 powers? Are there infinitely many terms whose sum consists of exactly 5 powers?
If it exists, a(15) > 10^20.

			a(1)  = 1422    = 2^5 + 2^7  + 5^3  + 5^4 + 2^9   = 5^2 + 7^2  + 3^5  + 4^5  + 9^2
a(2)  = 1464    = 2^5 + 2^6  + 2^7  + 4^5 + 6^3   = 5^2 + 6^2  + 7^2  + 5^4  + 3^6
a(3)  = 1554    = 2^3 + 2^7  + 8^2  + 5^4 + 3^6   = 3^2 + 7^2  + 2^8  + 4^5  + 6^3
a(4)  = 2612    = 2^5 + 2^6  + 5^3  + 7^3 + 2^11  = 5^2 + 6^2  + 3^5  + 3^7  + 11^2
a(5)  = 3127    = 2^3 + 2^9  + 6^3  + 7^3 + 2^11  = 3^2 + 9^2  + 3^6  + 3^7  + 11^2
a(6)  = 4481    = 2^6 + 7^2  + 2^10 + 2^11 + 6^4  = 6^2 + 2^7  + 10^2 + 11^2 + 4^6
a(7)  = 5644    = 4^5 + 9^2  + 10^2 + 7^3 + 4^6   = 5^4 + 2^9  + 2^10 + 3^7  + 6^4
a(8)  = 16122   = 2^3 + 4^3  + 2^8  + 5^6 + 13^2  = 3^2 + 3^4  + 8^2  + 6^5  + 2^13
a(9)  = 68521   = 2^8 + 4^5  + 4^6  + 3^10 + 8^4  = 8^2 + 5^4  + 6^4  + 10^3 + 4^8
a(10) = 77129   = 4^6 + 2^12 + 7^4  + 10^3 + 2^16 = 6^4 + 12^2 + 4^7  + 3^10 + 16^2
a(11) = 82583   = 2^5 + 4^3  + 12^2 + 7^5 + 2^16  = 5^2 + 3^4  + 2^12 + 5^7  + 16^2
a(12) = 1065585 = 2^9 + 2^12 + 7^4  + 10^4 + 2^20 = 9^2 + 12^2 + 4^7  + 4^10 + 20^2
a(13) = 4227140 = 5^6 + 13^2 + 7^4  + 11^4 + 2^22 = 6^5 + 2^13 + 4^7  + 4^11 + 22^2
a(14) = 6164560 = 5^7 + 2^18 + 9^5  + 21^2 + 7^8  = 7^5 + 18^2 + 5^9  + 2^21 + 8^7

Math StackExchange, Base-Exponent Invariants, 2020.
Matej Veselovac, PYTHON program for A337671
Eric Weisstein's World of Mathematics, Perfect Power.

Cf. A337670, A005188 (perfect digital invariants), perfect powers: A001597, A072103.

A337670 Numbers that can be expressed as both Sum x^y and Sum y^x where the x^y are not equal to y^x for any (x,y) pair and all (x,y) pairs are distinct.

Original entry on oeis.org

432, 592, 1017, 1040, 1150, 1358, 1388, 1418, 1422, 1464, 1554, 1612, 1632, 1713, 1763, 1873, 1889, 1966, 1968, 1973, 1990, 2091, 2114, 2190, 2291, 2320, 2364, 2451, 2589, 2591, 2612, 2689, 2697, 2719, 2753, 2775, 2803, 2813, 2883, 3087, 3127, 3141, 3146

Offset: 1

Matej Veselovac, Sep 15 2020

nonn

Numbers m of form m = Sum_{i=1...k} b_i^e_i = Sum_{i=1...k} e_i^b_i such that b_i^e_i != e_i^b_i, b_i > 1, e_i > 1, k = |{{b_i, e_i}, i = 1, 2, ...}|, k > 1.
Terms of the sequence relate to the Diophantine equation Sum_{i=1...k} x_i = 0, k > 1, x_i != 0, where x_i = (b_i^e_i - e_i^b_i) such that b_i > 1, e_i > 1 and (i != j) => ({b_i, e_i} != {b_j, e_j}). That is, we are observing linear combinations of elements from {(r^n - n^r) : n,r > 1} \ {0}, under given conditions.
For sums with k = 20 terms, one infinite family of examples is known: "2^(2t) + t^(4) + 2^(2t+8) + (t+4)^(4) + 2^(2t+16) + (t+8)^(4) + 2^(2t+32) + (t+16)^(4) + 2^(2t+34) + (t+17)^(4) + 4^(t+1) + (2t+2)^(2) + 4^(t+2) + (2t+4)^(2) + 4^(t+10) + (2t+20)^(2) + 4^(t+14) + (2t+28)^(2) + 4^(t+18) + (2t+36)^(2)" is a term of the sequence, for every t > 4.

			17 = 2^3 + 3^2 = 3^2 + 2^3 is not in the sequence because {2,3} = {3,2} are not distinct.
25 = 3^3 + 2^4 = 3^3 + 4^2 is not in the sequence because 3^3 = 3^3 and 2^4 = 4^2 are commutative.
The smallest term of the sequence is:
  a(1) = 432 = 3^2 + 5^2 + 2^6 + 3^4 + 5^3 + 2^7
             = 2^3 + 2^5 + 6^2 + 4^3 + 3^5 + 7^2.
The smallest term that has more than one representation is:
  a(11) = 1554 = 3^2 + 7^2 + 6^3 + 2^8 + 4^5
               = 2^3 + 2^7 + 3^6 + 8^2 + 5^4,
  a(11) = 1554 = 3^2 + 5^2 + 2^6 + 10^2 + 2^7 + 3^5 + 2^8 + 3^6
               = 2^3 + 2^5 + 6^2 + 2^10 + 7^2 + 5^3 + 8^2 + 6^3.
Smallest terms with k = 5, 6, 7, 8, 9, 10 summands are:
  a(9)  = 1422 = 5^2 + 7^2 + 9^2 + 3^5 + 4^5
               = 2^5 + 2^7 + 2^9 + 5^3 + 5^4,
  a(1)  = 432  = 3^2 + 5^2 + 2^6 + 3^4 + 5^3 + 2^7
               = 2^3 + 2^5 + 6^2 + 4^3 + 3^5 + 7^2,
  a(2)  = 592  = 3^2 + 5^2 + 7^2 + 4^3 + 2^6 + 5^3 + 2^8
               = 2^3 + 2^5 + 2^7 + 3^4 + 6^2 + 3^5 + 8^2,
  a(11) = 1554 = 3^2 + 5^2 + 2^6 + 10^2 + 2^7 + 3^5 + 2^8 + 3^6
               = 2^3 + 2^5 + 6^2 + 2^10 + 7^2 + 5^3 + 8^2 + 6^3,
  a(14) = 1713 = 3^2 + 2^5 + 6^2 + 8^2 + 4^3 + 2^7 + 3^5 + 2^9 + 5^4
               = 2^3 + 5^2 + 2^6 + 2^8 + 3^4 + 7^2 + 5^3 + 9^2 + 4^5,
  a(28) = 2451 = 3^2 + 5^2 + 6^2 + 8^2 + 3^4 + 2^7 + 6^3 + 3^5 + 5^4 + 2^10
               = 2^3 + 2^5 + 2^6 + 2^8 + 4^3 + 7^2 + 3^6 + 5^3 + 4^5 + 10^2.

Math StackExchange, Base-Exponent Invariants, 2020.
Matej Veselovac, PYTHON program for A337670
Eric Weisstein's World of Mathematics, Perfect Power.

Cf. A337671 (subsequence for k <= 5).
Cf. A005188 (perfect digital invariants).
Cf. Perfect powers: A001597, A072103.
Cf. Commutative powers: A271936.
Cf. Nonnegative numbers of the form (r^n - n^r), for n,r > 1: A045575.
Cf. Numbers of the form (r^n - n^r): A024012 (r = 2), A024026 (r = 3), A024040 (r = 4), A024054 (r = 5), A024068 (r = 6), A024082 (r = 7), A024096 (r = 8), A024110 (r = 9), A024124 (r = 10), A024138 (r = 11), A024152 (r = 12).

A337113 Triangle read by rows in which row n lists all prime factors of A337112(n) in increasing order.

Original entry on oeis.org

0, 2, 2, 0, 0, 0, 2, 3, 3, 5, 3, 3, 3, 5, 5, 2, 3, 3, 3, 3, 7, 2, 3, 3, 3, 3, 3, 13, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 13, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 17, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 43, 43

Offset: 1

Matej Veselovac, Aug 16 2020

If A337112(n) = 0, then n 0's are listed instead.

			A337112(1)  = 0.
A337112(2)  = 2*2.
A337112(3)  = 0*0*0.
A337112(4)  = 2*3*3*5.
A337112(5)  = 3*3*3*5*5.
A337112(6)  = 2*3*3*3*3*7.
A337112(7)  = 2*3*3*3*3*3*13.
A337112(8)  = 2*3*3*3*3*3*3*3.
A337112(9)  = 3*3*3*3*3*3*3*3*5.
A337112(10) = 3*3*3*3*3*3*3*3*3*13.
A337112(11) = 3*3*3*3*3*3*3*3*3*3*17.
A337112(12) = 3*3*3*3*3*3*3*3*3*3*3*3.
A337112(13) = 3*3*3*3*3*3*3*3*3*3*3*43*43.

Cf. A337112 (products of rows).
Cf. A337037, A337080, A337081.
Cf. A056472 (all factorizations of n).

A337112 Smallest term of A337081 that has exactly n prime factors, or 0 if no such term exists.

Original entry on oeis.org

0, 4, 0, 90, 675, 1134, 6318, 4374, 32805, 255879, 1003833, 531441, 327544803, 20751953125, 225830078125, 91552734375, 1068115234375, 23651123046875, 316619873046875, 1697540283203125, 13256072998046875, 85353851318359375, 541210174560546875, 4518032073974609375, 58233737945556640625

Offset: 1

Matej Veselovac, Aug 16 2020

nonn

a(n) is the smallest product of n primes that has unordered factorizations whose sums of factors are the same (is a term of A337080) and all of whose proper divisors have the complementary property: that every unordered factorization has a distinct sum of factors (i.e., all proper divisors are terms of A337037).

Cf. A337113 (factors of terms).
Cf. A337037, A337080, A337081.
Cf. A056472 (all factorizations of n).
Cf. r-almost primes: A000040 (r = 1), A001358 (r = 2), A014612 (r = 3), A014613 (r = 4), A014614 (r = 5), A046306 (r = 6), A046308 (r = 7), A046310 (r = 8), A046312 (r = 9), A046314 (r = 10), A069272 (r = 11), A069273 (r = 12), A069274 (r = 13), A069275 (r = 14), A069276 (r = 15), A069277 (r = 16), A069278 (r = 17), A069279 (r = 18), A069280 (r = 19), A069281 (r = 20).

a(14) onward from David A. Corneth, Aug 26 2020

A337037 Numbers whose every unordered factorization has a distinct sum of factors.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25, 26, 27, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 45, 46, 47, 49, 50, 51, 53, 54, 55, 57, 58, 59, 61, 62, 63, 65, 66, 67, 69, 70, 71, 73, 74, 75, 77, 78, 79, 81, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 98, 99, 101

Offset: 1

Matej Veselovac, Aug 12 2020

The number 1 is in the sequence by convention.
All primes p are trivially in the sequence.
All semiprimes greater than 4 are in the sequence because they have only two unordered factorizations pq = p*q whose sums are distinct. They are distinct because the only solution to p*q = p+q is p=q=2.
If a number m is not in the sequence, then all multiples of m are not in the sequence. For example, multiples of 4 are not in the sequence because there always exist at least two factorizations 4*k = 2*2*k whose factors sum to the same value 4+k = 2+2+k.
The complement is in A337080.
Numbers m such that A069016(m) = A001055(m). - Michel Marcus, Aug 15 2020

			All unordered factorization of 30 are 30 = 2*15 = 3*10 = 5*6 = 2*3*5. Corresponding sums of factors are distinct: 30, 17 = 15+2, 13 = 10+3, 11 = 6+5, 10 = 2+3+5. Therefore 30 is in the sequence.
All unordered factorization of 90 are 90 = 45*2 = 30*3 = 18*5 = 15*6 = 15*3*2 = 10*9 = 9*5*2 = 10*3*3 = 6*5*3 = 5*3*3*2. Corresponding sums of factors are not all distinct: 90, 57, 33, 23, 21, 20, 19, 16, 16, 14, 13 because the sum 16 = 10+3+3 = 9+5+2 appears twice. Therefore 90 is not in the sequence.

Eric Weisstein's World of Mathematics, Unordered Factorization.

Cf. A337080 (complement), A337081 (primitive complement).
Cf. A001055 (number of unordered factorizations of n), A074206 (number of ordered factorizations of n).
Cf. A056472 (all factorizations of n), A069016 (number of distinct sums).

factz(n, minn) = {my(v=[]); fordiv(n, d, if ((d>=minn) && (d<=sqrtint(n)), w = factz(n/d, d); for (i=1, #w, w[i] = concat([d], w[i]);); v = concat(v, w););); concat(v, [[n]]);}
factorz(n) = factz(n, 2);
isok(n) = my(vs = apply(x->vecsum(x), factorz(n))); #vs == #Set(vs); \\ Michel Marcus, Aug 13 2020

A337081 Primitive complement of A337037: terms of A337080 that are not multiples of previous terms.

Original entry on oeis.org

4, 90, 546, 675, 850, 918, 945, 1026, 1050, 1134, 1242, 1365, 1386, 1575, 1650, 1750, 1782, 1950, 2205, 2295, 2310, 2450, 2475, 2646, 2793, 2850, 3250, 3366, 3465, 3626, 3654, 3762, 3850, 3969, 3990, 4218, 4290, 4374, 4455, 4510, 4550, 4650, 4875, 4998, 5022, 5166, 5382, 5390, 5610

Offset: 1

Matej Veselovac, Aug 14 2020

The only semiprime in the sequence is a(1) = 4, and there are no terms with exactly 3 prime factors.

Numbers of form p^k where p >= 5 is a prime number are terms of the sequence if and only if k = 4p+6. The only terms of the form 2^k or 3^k have k = 2, 12 respectively.

			Numbers of the form m = 2*p*q*((p-1)*q-(p-2)) where p, q and (p-1)*q-(p-2) are odd prime numbers are even terms of the sequence. First, notice that m is a term of A337080 because the factorizations m = (2*((p-1)*q-(p-2)))*(p)*(q) = (2)*(((p-1)*q-(p-2)))*(p*q) have equal sums of factors. Second, m is not a multiple of any of the previous terms of the sequence because m has exactly 4 prime factors and the only term with less than 4 prime factors is 4, but 4 does not divide m.

David A. Corneth, Table of n, a(n) for n = 1..10000
Math StackExchange, Smallest power of a prime whose factorizations don't have distinct sums of factors, 2020.

Cf. A337037, A337080, A337112 (smallest term with n factors).
Cf. A001055 (number of unordered factorizations of n), A074206 (number of ordered factorizations of n).
Cf. A056472 (all factorizations of n), A069016 (number of distinct sums).

factz(n, minn) = {my(v=[]); fordiv(n, d, if ((d>=minn) && (d<=sqrtint(n)), w = factz(n/d, d); for (i=1, #w, w[i] = concat([d], w[i]);); v = concat(v, w););); concat(v, [[n]]);}
factorz(n) = factz(n, 2);
isok(n) = my(vs = apply(x->vecsum(x), factorz(n))); #vs != #Set(vs);
isprimitive(n, va) = {for (k=1, #va, if ((n % va[k]) == 0, return (0));); return (1);}
lista(nn) = {my(va = []); for (n=1, nn, if (isok(n) && isprimitive(n, va), va = concat(va, n));); va;} \\ Michel Marcus, Aug 15 2020

A337080 Complement of A337037.

Original entry on oeis.org

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 90, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232, 236, 240, 244, 248

Offset: 1

Matej Veselovac, Aug 14 2020

Numbers with a pair of unordered factorizations whose sums of factors are the same.
All terms of the sequence are composite.
The smallest odd term of the sequence is a(174) = 675. This is a term of the sequence because 675 = 27*5*5 = 9*3*25 and 27+5+5 = 9+3+25 = 37.
Terms of the sequence are used in variations of a logic puzzle known as "Ages of Three Children Puzzle" or "Census-taker problem". For the original puzzle, see A334911.
If a number m is in the sequence, then all multiples of m are in the sequence. For example, multiples of 4 are in the sequence because there always exist at least two factorizations 4*k = 2*2*k whose factors sum to the same value 4+k = 2+2+k.
Numbers m such that A069016(m) < A001055(m). - Michel Marcus, Aug 15 2020

			All unordered factorization of 90 are 90 = 45*2 = 30*3 = 18*5 = 15*6 = 15*3*2 = 10*9 = 9*5*2 = 10*3*3 = 6*5*3 = 5*3*3*2. Corresponding sums of factors are not all distinct: 90, 57, 33, 23, 21, 20, 19, 16, 16, 14, 13 because the sum 16 = 10+3+3 = 9+5+2 appears twice. Therefore 90 is in the sequence.
All unordered factorization of 30 are 30 = 15*2 = 10*3 = 6*5 = 5*3*2. Corresponding sums of factors are all distinct: 30 = 30, 17 = 15+2, 13 = 10+3, 11 = 6+5, 10 = 2+3+5. Therefore 30 is not in the sequence.

Eric Weisstein's World of Mathematics, Unordered Factorization.

Cf. A334911 (census-taker numbers).
Cf. A337037 (complement), A337081.
Cf. A001055 (number of unordered factorizations of n), A074206 (number of ordered factorizations of n).
Cf. A056472 (all factorizations of n), A069016 (number of distinct sums).

factz(n, minn) = {my(v=[]); fordiv(n, d, if ((d>=minn) && (d<=sqrtint(n)), w = factz(n/d, d); for (i=1, #w, w[i] = concat([d], w[i]);); v = concat(v, w););); concat(v, [[n]]);}
factorz(n) = factz(n, 2);
isok(n) = my(vs = apply(x->vecsum(x), factorz(n))); #vs != #Set(vs); \\ Michel Marcus, Aug 14 2020

Edited by N. J. A. Sloane, Sep 14 2020

A334917 Indices in A334916 where records occur.

Original entry on oeis.org

4, 5, 6, 7, 8, 10, 18, 24, 32, 43, 73

Offset: 1

Matej Veselovac, May 16 2020

a(n) cannot be a perfect square if n>1.

If the conjecture "A334916(n)>0 for n>3" is true, then the next term is a(12)=107.

Cf. A000290 (perfect squares), A334916.

A334916 a(n) is the smallest number > 1 whose base n digits yield the original number when added and multiplied left to right; or 0 if no such number exists.

Original entry on oeis.org

0, 0, 0, 6, 12, 160, 324, 405, 12, 8385, 36, 189, 784, 32, 1656, 20, 721, 25215, 80, 45, 559, 2585, 5525, 323844, 30, 160, 60, 90, 150, 1071, 11650, 1038448, 6275, 2669, 77, 42, 2224, 324224, 1817, 2016, 252, 7425, 1593074855, 96, 5450, 192, 345906, 23541, 56

Offset: 1

Matej Veselovac, May 16 2020

These numbers have been called "baseless in base n".
a(n) is divisible by its last base n digit.
The number 8385 = ((((8)8+3)3+8)8+5)5 is known to be the unique baseless number in base 10. Are there number bases n, other than 6 and 10, that have a unique example?
If the term a(107) is not zero, then it is at least a(107) > 107^6 > 1.5*10^12. Is it true that a(n)>0 for all n>3?

			Every number can be written as A = (...((((a)N+b)N+c)N+d)...) where a,b,c,d,... are digits of number A in base N. If we take that expression and replace the "multiplications by base N" with "multiplications by digits a,b,c,d,..." and also multiply it with the last digit to use up all digits, we get some number A*. If it holds A = A*, then we say number A is a baseless number.
For example, the decimal number base has only one baseless number:
.
a(10) = 8385 = ((((8)*10+3)*10+8)*10+5) = ((((8)*8+3)*3+8)*8+5)*5.
.
There are at most finitely many baseless numbers for every fixed number base. For example, the number base 4 has exactly three baseless numbers:
.
6 = ((1)*4+2)        = ((1)*1+2)*2       = 12_4;
27 = (((1)*4+2)*4+3) = (((1)*1+2)*2+3)*3 = 123_4;
46 = (((2)*4+3)*4+2) = (((2)*2+3)*3+2)*2 = 232_4;
.
The smallest of them is 6, hence a(4)=6.

Math StackExchange user "Vepir" (Matej Veselovac), Terms a(n) < 10^10 for n < 500, and including the 11th record a(73) ~ 2*10^10.
Math StackExchange, Does every number base have at least one "baseless number"?

Cf. A000290 (perfect squares), A334917 (indices of records).

\\ for n>=4
isok(k,n) = {my(d=digits(k, n), s=0); for (i=1, #d, s = (s+d[i])*d[i];); s == k;}
a(n) = {my(k=2); while (!isok(k, n), k++); k;} \\ Michel Marcus, Jun 18 2020

If n is a perfect square, then a(n) = n + sqrt(n). Otherwise, a(n) > 2n.

A333837 Number of ways to collapse an n-rowed triangular formation of dominoes.

Original entry on oeis.org

1, 2, 5, 18, 97, 802, 10565, 228850, 8289217, 506526530, 52501381765, 9260170733266, 2784551512218145, 1429063630276963426, 1252517782851235507141, 1875484239084442842046130, 4798818821638537354534159233, 20984654018757393270224583817858

Offset: 0

Matej Veselovac, Apr 07 2020

Every domino either falls or does not. A domino can fall if and only if one of the two dominoes above it falls. The total number of dominoes in the formation is A000217(n).

			A domino can fall if and only if one of the two dominoes above it falls.
For n=1,2,3,4,... we have the following domino formations:
                               0
                    0         0 0
           0       0 0       0 0 0
    0,    0 0,    0 0 0,    0 0 0 0,  ...
For n=1, we have a single domino which either falls or does not, hence a(1)=2.
For n=2, we have two rows of dominoes in the triangular formation: If the top domino falls, then the bottom two can both fall (or not) giving 2^2=4 scenarios. The fifth scenario is when the first domino does not fall, then the bottom two can't either. This gives a(2)=4+1=5.
For n=3, we have three rows of dominoes in the triangular formation: If both dominoes in the second row fall, then the third row can fall in 2^3=8 ways. If only one of the two dominoes in the second row falls, then the third row can fall in 2^2 ways in both cases which totals 2*2^2=8. If none of the two dominoes in the second row fall, then we know the top domino is in either of the 2 states. Summing this up gives a(3)=18.
For n=4, we have four rows of dominoes in the triangular formation: We have a(4) = 18 + 7*(2^2-1) + 4*(2^3-1) + 2*(2^4-1) = 97, because there are 7,4,2 ways to have 2,3,4 potentially falling dominoes in the last row, and 18 is the number of ways for the previous rows to fall if we ignore the last row. We subtract 1 state from every 2^k states of k potentially falling dominoes, because those states are counted in the previous added 18 scenarios.

Math StackExchange user Bartek, Table of n, a(n) for n = 0..26
Math StackExchange user Vepir, Domino matrix M, for n=10
Math StackExchange, A sequence with dominoes

Cf. A000217 (triangular numbers).

a(n) = a(n-1) + Sum_{k=2..n} b(n,k)*(2^k-1), where b(n,k) is the number of ways to have k potentially falling dominoes in the n-th row of the domino triangle. We know b(n,2) = Sum_{k=2..n-1} k*b(n-1,k), but b(n,k) for k>=3 does not appear to have a simple recursion, and needs to be calculated explicitly, step by step.
a(n) = norm1(v(n)), where v(n) = v(n-1)*M is a sequence of (2^n+1)-dimensional vectors starting with v(1) = e1 + e2 = (1,1,0,...). The matrix M is the (2^n+1)x(2^n+1) dimensional "domino matrix" M = ((1,0,...),(1,1,1,1,0,...),(1,0,1,0,1,0,1,0,...),(1,1,1,1,1,1,1,1,0,...),...) which is calculated as follows: The M(i,j) entry (where i,j = 0,1,2,...) of the matrix M is 1 if and only if the binary representations of i,j are in a valid state (0's are standing dominoes and 1's are fallen dominoes) when taken as consecutive rows of the triangular domino formation, otherwise it is 0. For example, i=2 has the binary representation 01. If we look at the next row of the triangular domino formation, then the following formations are possible:
     1 0     1 0     1 0     1 0
    0 0 0,  0 1 0,  1 0 0,  1 1 0. Translating the next row scenarios j = 000, 010, 100, 110 from binary to decimal, we obtain j = 0, 2, 4, 6. Hence, the i=2 row of the domino matrix M will start as (1,0,1,0,1,0,1). The remainder of the row is filled with zeros.

a(23)-a(26) from Bartlomiej Bollin, Apr 14 2020

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User: Matej Veselovac

A337671 Subsequence of A337670 in which there are at most five terms in the sum.

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A337670 Numbers that can be expressed as both Sum x^y and Sum y^x where the x^y are not equal to y^x for any (x,y) pair and all (x,y) pairs are distinct.

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A337113 Triangle read by rows in which row n lists all prime factors of A337112(n) in increasing order.

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A337112 Smallest term of A337081 that has exactly n prime factors, or 0 if no such term exists.

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A337037 Numbers whose every unordered factorization has a distinct sum of factors.

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A337080 Complement of A337037.

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A334917 Indices in A334916 where records occur.

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