Giovanni Corbelli - cp's OEIS frontend

A351374 Base-20 Armstrong or narcissistic numbers (written in base 10).

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2413, 53808, 760400, 760401, 45661018, 62470211, 619939142, 14613048357, 1421043363262183, 48470736648305918, 514822672411130775, 360672575087017687943, 264237343348909655564587, 267218514330351511200145

Offset: 1

Giovanni Corbelli, Mar 18 2022

Written in base twenty the numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, 60D, 6EA8, 4F100, 4F101, E57CAI, JA8FAB, 9DEC7H2, B86BB0HH.

Sequence is finite. Since k*19^k < 20^(k-1) for k >= 157, all terms must have less than 157 base-20 digits. 20*m is a term if and only if 20*m+1 is a term. - Chai Wah Wu, Apr 20 2022

			2413 is in the sequence because 2413 is 60D in base 20 (D stands for 13) and 6^3 + 0^3 + 13^3 = 2413. (The exponent 3 is the number of base-20 digits.)

Jinyuan Wang, C program
Jinyuan Wang, Table of base-20 Armstrong numbers
Eric Weisstein's World of Mathematics, Narcissistic Number

In other bases: A010344 (base 4), A010346 (base 5), A010348 (base 6), A010350 (base 7), A010354 (base 8), A010353 (base 9), A005188 (base 10), A161948 (base 11), A161949 (base 12), A161950 (base 13), A161951 (base 14), A161952 (base 15), A161953 (base 16).

Select[Range[10^6], # == Total[ IntegerDigits[#, 20]^IntegerLength[#, 20]] &]

isok(m) = my(d=digits(m, 20)); sum(k=1, #d, d[k]^#d) == m; \\ Michel Marcus, Mar 19 2022

from itertools import islice, combinations_with_replacement
from sympy.ntheory.factor_ import digits
def A351374_gen(): # generator of terms
    for k in range(1,157):
        a = tuple(i**k for i in range(20))
        yield from (x[0] for x in sorted(filter(lambda x:x[0] > 0 and tuple(sorted(digits(x[0],20)[1:])) == x[1], \
                          ((sum(map(lambda y:a[y],b)),b) for b in combinations_with_replacement(range(20),k)))))
A351374_list = list(islice(A351374_gen(),20)) # Chai Wah Wu, Apr 20 2022

a(28)-a(30) from Chai Wah Wu, Apr 20 2022

a(31)-a(33) from Jinyuan Wang, May 05 2025

A349296 First differences of A349295.

Original entry on oeis.org

1, 14, 109, 479, 1570, 4031, 8997, 17948, 32853, 56408, 91776, 143003, 215196, 313732, 444813, 616816, 839685, 1120435, 1472736, 1907995, 2440463, 3086644, 3861599, 4784197, 5878808, 7160841, 8659826, 10399512, 12407231, 14710254, 17351756

Offset: 1

Giovanni Corbelli, Nov 13 2021

nonn

a(n) is the number of ordered 6-tuples (a_1,a_2,a_3,a_4,a_5,a_6) having all terms in {1,...,n}, with at least one element equal to n, such that there exists a tetrahedron ABCD with those edge-lengths, taken in a particular order (see comments in A349295).

Conjecture: for n tending to infinity the ratio a(n) / A097125(n) tends to 24 as the probability that all a_i's are different tends to 1 and there are 24 6-tuples corresponding to the same tetrahedron if all a_i's are different. For n=254 the ratio is 23.9936919.

Giovanni Corbelli, Table of n, a(n) for n = 1..254
Sascha Kurz, Enumeration of integral tetrahedra, arXiv:0804.1310 [math.CO], 2008.

Cf. A097125, A349295.

A349295 a(n) is the number of ordered 6-tuples (a_1,a_2,a_3,a_4,a_5,a_6) having all terms in {1,...,n} such that there exists a tetrahedron ABCD with those edge-lengths, taken in a particular order (see comments).

Original entry on oeis.org

0, 1, 15, 124, 603, 2173, 6204, 15201, 33149, 66002, 122410, 214186, 357189, 572385, 886117, 1330930, 1947746, 2787431, 3907866, 5380602, 7288597, 9729060, 12815704, 16677303, 21461500, 27340308, 34501149, 43160975, 53560487, 65967718, 80677972, 98029728

Offset: 0

Giovanni Corbelli, Nov 13 2021

nonn

Edges with length a_1,a_2,a_3 form a face, a_1 is opposite to a_4, a_2 is opposite to a_5, a_3 is opposite to a_6. If the a_i's are all different, then there are 24 6-tuples corresponding to the same tetrahedron. The tetrahedron is possible iff triangular inequalities hold for every face and the Cayley-Menger determinant is positive. It has been proved that if triangular inequalities hold for at least one face and the Cayley-Menger determinant is positive, then the triangular inequalities for the other three faces hold, too (see article by Wirth, Dreiding in links, (5) at page 165).

Conjecture: The ratio a(n)/n^6 decreases with n and tends to a limit which is 0.10292439+-0,00000024 (1.96 sigmas, 95% confidence level) evaluated for n=2^32 on 6.4*10^12 random 6-tuples.

			For n=2 the 6-tuples are
(1,1,1,1,1,1),
(1,1,1,2,2,2), (1,2,2,2,1,1), (2,1,2,1,2,1), (2,2,1,1,1,2),
(2,2,1,2,2,1), (2,1,2,2,1,2), (1,2,2,1,2,2),
(1,2,2,2,2,2), (2,1,2,2,2,2), (2,2,1,2,2,2), (2,2,2,1,2,2), (2,2,2,2,1,2), (2,2,2,2,2,1),
(2,2,2,2,2,2)
corresponding to A097125(1) + A097125(2) = 5 different tetrahedra.

Giovanni Corbelli, Table of n, a(n) for n = 0..254
Giovanni Corbelli, FreeBasic program
Karl Wirth and André S. Dreiding, Edge lengths determining tetrahedrons, Elemente der Mathematik, Volume 64, Issue 4, 2009, pp. 160-170.

Cf. A097125, A349296, A346575.

A346638 a(n) is the number of 6-tuples (a_1,a_2,a_3,a_4,a_5,a_6) having all terms in {1,...,n} such that there exists a hexagon with these side-lengths.

Original entry on oeis.org

0, 1, 64, 729, 4096, 15619, 46614, 117481, 261640, 530181, 997228, 1766017, 2975688, 4808791, 7499506, 11342577, 16702960, 24026185, 33849432, 46813321, 63674416, 85318443, 112774222, 147228313, 190040376, 242759245, 307139716, 385160049, 479040184, 591260671

Offset: 0

Giovanni Corbelli, Jul 26 2021

The existence of such a six-sided polygon implies that every element of the sextuple is less than the sum of the other elements.

Giovanni Corbelli, Visual Basic routine for generating number of six-sided polygons
Giovanni Corbelli Proof of the formula: Number of k-tuples with elements in {1,2,...,N} corresponding to k-sided polygons
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A006003, A346636, A346637.

CoefficientList[Series[x (1+57x+302x^2+302x^3+51x^4+x^5)/(1-x)^7,{x,0,40}],x] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,1,64,729,4096,15619,46614},40] (* Harvey P. Dale, Oct 30 2024 *)

a(n) = n^6 - 6*binomial(n+1,6) = n^6 - (n+1)*binomial(n,5).
General formula for k-tuples: a_k(n) = n^k - k*binomial(n+1,k) = n^k - (n+1)*binomial(n,k-1).
G.f.: x*(1 + 57*x + 302*x^2 + 302*x^3 + 51*x^4 + x^5)/(1 - x)^7. - Stefano Spezia, Sep 27 2021

A346637 a(n) is the number of quintuples (a_1,a_2,a_3,a_4,a_5) having all terms in {1,...,n} such that there exists a pentagon with these side-lengths.

Original entry on oeis.org

0, 1, 32, 243, 1019, 3095, 7671, 16527, 32138, 57789, 97690, 157091, 242397, 361283, 522809, 737535, 1017636, 1377017, 1831428, 2398579, 3098255, 3952431, 4985387, 6223823, 7696974, 9436725, 11477726, 13857507, 16616593, 19798619, 23450445, 27622271, 32367752

Offset: 0

Giovanni Corbelli, Jul 26 2021

The existence of such a five-sided polygon implies that every element of the quintuple is less than the sum of the other elements.

Giovanni Corbelli, Visual Basic routine for generating number of five-sided polygons
Giovanni Corbelli Proof of the formula: Number of k-tuples with elements in {1,2,...,N} corresponding to k-sided polygons
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A006003, A346636, A346638.

a(n) = n^5 - 5*binomial(n+1,5) = n^5 - (n+1)*binomial(n,4).
General formula for k-tuples: a_k(n) = n^k - k*binomial(n+1,k) = n^k - (n+1)*binomial(n,k-1).
G.f.: x*(1 + 26*x + 66*x^2 + 21*x^3 + x^4)/(1 - x)^6. - Stefano Spezia, Sep 27 2021

A346636 a(n) is the number of quadruples (a_1, a_2, a_3, a_4) having all terms in {1,...,n} such that there exists a quadrilateral with these side lengths.

Original entry on oeis.org

0, 1, 16, 77, 236, 565, 1156, 2121, 3592, 5721, 8680, 12661, 17876, 24557, 32956, 43345, 56016, 71281, 89472, 110941, 136060, 165221, 198836, 237337, 281176, 330825, 386776, 449541, 519652, 597661, 684140, 779681, 884896, 1000417, 1126896, 1265005, 1415436

Offset: 0

Giovanni Corbelli, Jul 26 2021

nonn

The existence of such a four-sided polygon implies that every element of the quadruple is less than the sum of the other elements.

Giovanni Corbelli, Visual Basic routine for generating number of four-sided polygons
Giovanni Corbelli Proof of the formula: Number of k-tuples with elements in {1,2,…,N} corresponding to k-sided polygons
Sean A. Irvine, Java program (github)

Cf. A006003, A346637, A346638.

Formula: a(n) = n^4 - 4*binomial(n+1,4) = n^4 - (n+1)*binomial(n,3).

General formula for k-tuples: a_k(n) = n^k - k*binomial(n+1,k) = n^k - (n+1)*binomial(n,k-1).

A346575 a(n) is the number of 6-tuples (a_1,a_2,a_3,a_4,a_5,a_6) having all terms in {1,...,n} such that there exists a tetrahedron ABCD with those edge-lengths.

Original entry on oeis.org

0, 1, 43, 327, 1792, 6139, 17607, 43291, 96142, 193149, 362383, 638533, 1075110, 1733023, 2700217, 4076133, 5994310, 8611819, 12119139, 16738861, 22746004, 30449013, 40212679, 52452031, 67651170, 86348035, 109166881, 136796079, 170024038, 209707144, 256814946, 312433795

Offset: 0

Giovanni Corbelli, Jul 24 2021

nonn

The existence of such a tetrahedron implies the following:
(1) there exists at least one permutation (a_i1,a_i2,a_i3,a_i4,a_i5,a_i6) such that triangular inequalities hold for (a_i1,a_i2,a_i3) (BCD), (a_i1,a_i4,a_i5) (ABC), (a_i2,a_i5,a_i6) (ACD) and (a_i3,a_i6,a_i4) (ABD), where we have a_i1=BC, a_i2=CD, a_i3=DB, a_i4=AB, a_i5=AC, a_i6=AD;
(2) a tetrahedron with such edge-lengths can be built.
Values were computed using a Visual Basic program with two different routines, manually checked for n = 2 and n = 3.
Conjecture 1: a(n)/n^6 tends to a limit which is 0.338170 +- 0.000017 (confidence level 95%). This number has been evaluated with a Monte-Carlo test on 3 billion sextuples with random values in (0,1) which simulate n -> oo.
Conjecture 2: there is no polynomial formula for a(n), as finite difference method fails.

			For a(2)=43 the solutions are (1,1,1,1,1,1), all 20 permutations of (1,1,1,2,2,2), all 15 permutations of (1,1,2,2,2,2), all 6 permutations of (1,2,2,2,2,2) and (2,2,2,2,2,2).

Chai Wah Wu, Table of n, a(n) for n = 0..52
Lucas A. Brown, Python program.
Giovanni Corbelli, Visual Basic routine generating number of tetrahedra.
Karl Wirth and Andre Dreiding, Edge lengths determining tetrahedrons, Elemente der Mathematik, 64 (2009), 160-170.

Cf. A097125.

Equivalent sequence for triples with respect to triangles: A006003.

Python
```
# See LINKS.
```

Conjecture: Limit_{n->oo} a(n)/n^6 exists and is approximately 0.33817.

a(21)-a(31) from Lucas A. Brown, Mar 13 2024

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User: Giovanni Corbelli

A351374 Base-20 Armstrong or narcissistic numbers (written in base 10).

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A349296 First differences of A349295.

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A349295 a(n) is the number of ordered 6-tuples (a_1,a_2,a_3,a_4,a_5,a_6) having all terms in {1,...,n} such that there exists a tetrahedron ABCD with those edge-lengths, taken in a particular order (see comments).

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A346638 a(n) is the number of 6-tuples (a_1,a_2,a_3,a_4,a_5,a_6) having all terms in {1,...,n} such that there exists a hexagon with these side-lengths.

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A346637 a(n) is the number of quintuples (a_1,a_2,a_3,a_4,a_5) having all terms in {1,...,n} such that there exists a pentagon with these side-lengths.

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A346636 a(n) is the number of quadruples (a_1, a_2, a_3, a_4) having all terms in {1,...,n} such that there exists a quadrilateral with these side lengths.

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A346575 a(n) is the number of 6-tuples (a_1,a_2,a_3,a_4,a_5,a_6) having all terms in {1,...,n} such that there exists a tetrahedron ABCD with those edge-lengths.

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